\ 


•s-nV 


*-'''^^.. 


University  of  California  •  Berkeley 


The  Theodore  P.  Hill  Collection 

of 
Early  American  Mathematics  Books 


THE 


ELEMENTS  OF  EUCLID, 


VIZ. 


THE  FIRST  SIX  BOOKS, 


TOGETHER  WITH 


THE  ELEVENTH  AND  TWELFTH. 


THE  ERBORS 


BY  WHICH 


THEOJT,    OR  OTHERS,    HATE   LOXG   AGO    VITIATED   THESE    BOOKS    ARE    CORRECTED, 
AND   SOME    OF   EUCLID's   DEMONSTRATIONS    ARE    RESTORED. 


ALSO 


IN  LIKE  MANNER  CORRECTED. 


BY  ROBERT  SLMSON,  M.  D. 

EMERITUS  PROFESSOR  OF  MATHEMATICS  IN  THE  CNIVERSITT  OF  GLASGOW. 


TO  THIS  EDITION  ARE  ALSO  ANNEXED, 

ELEMENTS  OF  PLANE  AND  SPHERICAL  TRIGONOMETRY. 


Pttlatrel  jiiita : 

DESILVER,  Jr.  k  THOMAS,  247  MARKET  STREET 

John  C.  Clark,  Printer. 

1834. 


PREFACE. 


The  opinions  of  the  Moderns  concerning  the  author 
of  the  Elements  of  Geometry,  which  go  under  Euchd's 
name,  are  very  different,  and  contrary  to  one  another. 
Peter  Ramus  ascribes  the  Propositions,  as  well  as  their 
Demonstrations,  to  Theon;  others  think  the  Proposi- 
tions to  be  Euclid's,  but  that  the  Demonstrations  are 
Theon's;  and  others  maintain,  that  all  the  Propositions 
and  their  Demonstrations  are  Euclid's  own.  John 
Buteo  and  Sir  Henry  Savile  are  the  authors  of  great- 
est note  who  assert  this  last,  and  the  greater  part  of 
geometers  have  ever  since  been  of  this  opinion,  as  they 
thought  it  the  most  probable.  Sir  Henry  Savile,  after 
the  several  arguments  he  brings  to  prove  it,  makes  this 
conclusion  (page  13,  Prselect.)  "  That,  excepting  a 
very  few  interpolations,  explications,  and  additions, 
Theon  altered  nothing  in  Euclid."  But,  by  often  con- 
sidering and  comparing  together  the  Definitions  and 
Demonstrations  as  they  are  in  the  Greek  editions  we 
now  have,  I  found  that  Theon,  or  whoever  was  the 
editor  of  the  present  Greek  text,  by  adding  some 
things,  suppressing  others,  and  mixing  his  own  with 
Euclid's  Demonstrations,  had  changed  more  things  to 
the  worse  than  is  commonly  supposed,  and  those  not 
of  small  moment,  especially  in  the  fifth  and  eleventh 
Books  of  the  Elements,  which  this  editor  has  greatly 
vitiated;  for  instance,  by  substituting  a  shorter,  but 


IV  PREFACE. 


insufficient  Demonstration  of  the  18th  Prop,  of  the  5th 
Book,  in  place  of  the  legitimate  one  which  Euclid  had 
given;  and  by  taking  out  of  this  Book,  besides  other 
things,  the  good  definition  which  Eudoxus  or  Euclid 
had  given  of  compound  ratio,  and  giving  an  absurd  one 
in  place  of  it,  in  the  5th  Definition  of  the  6th  Book, 
whi'ich  neither  Euclid,  Archimedes,  Appollonius,  nor 
any  geometer  before  Theon's  time,  ever  made  use  of, 
and  of  which  there  is  not  to  be  found  the  least  appear- 
ance in  any  of  their  writings;  and,  as  this  Definition 
did  much  embarrass  beginners,  and  is  quite  useless,  it 
is  now  thrown  out  of  the  Elements,  and  another,  which, 
without  doubt,  Euclid  had  given,  is  put  in  its  proper 
place  among  the  Definitions  of  the  5th  Book,  by  which 
the  doctrine  of  compound  ratios  is  rendered  plain  and 
easy.  Besides,  among  the  Definitions  of  the  11th 
Book,  there  is  this,  which  is  the  10th,  viz.  "  Equal  and 
similar  solid  figures  are  those  which  are  contained  by 
similar  planes  of  the  same  number  and  magnitude." 
Now  this  Proposition  is  a  Theorem,  not  a  Definition; 
because  the  equality  of  figures  of  any  kind  must  be  de- 
monstrated, and  not  assumed;  and  therefore,  though 
this  were  a  true  Proposition,  it  ought  to  have  been 
demonstrated.  But,  indeed,  this  Proposition,  which 
makes  the  10th  Definition  of  the  11th  Book,  is  not 
true  universally,  except  in  the  case  in  which  each  of 
the  solid  angles  of  the  figures  is  contained  by  no  more 
than  three  plane  angles;  for  in  other  cases,  two  solid 
figures  may  be  contained  by  similar  planes  of  the  same 
number  and  magnitude,  and  yet  be  unequal  to  one  ano- 
ther, as  shall  be  made  evident  in  the  Notes  subjoined 
to  these  Elements.  In  like  manner,  in  the  Demonstra- 
tion of  the  26th  Prop,  of  the  11th  Book,  it  is  taken  for 
granted,  that  those  solid  angles  are  equal  to  one  ano- 


PREFACE. 


ther,  which  are  contained  by  plane  angles  of  the  same 
number  and  magnitude,  placed  in  the  same  order;  but 
neither  is  this  universally  true,  except  in  the  case  in 
which  the  solid  angles  are  contained  by  no  more  than 
three  plane  angles;  nor  of  this  case  is  there  any  De- 
monstration in  the  Elements  we  now  have,  though  it 
is  quite  necessary  there  should  be  one.  Now,  upon 
the  10th  Definition  of  this  Book  depend  the  25th  and 
28th  Propositions  of  it;  and  upon  the  25th  and  26th 
depend  other  eight,  viz.,  the  27th,  31st,  32d,  33d,  34th, 
36th,  37th,  and  40th,  of  the  same  Book;  and  the  12th 
of  the  12th  Book  depends  upon  the  8th  of  the  same; 
and  this  8th,  and  the  Corollary  of  Proposition  17th 
and  Proposition  18th  of  the  12th  Book,  depend  upon 
the  9th  Definition  of  the  11th  Book,  which  is  not  a 
right  definition,  because  there  may  be  solids  contained 
by  the  same  number  of  similar  plane  figures,  which  are 
not  similar  to  one  another,  in  the  true  sense  of  simi- 
larity received  by  all  geometers;  and  all  these  Propo- 
sitions have,  for  these  reasons,  been  insufficiently  de- 
monstrated since  Theon's  time  hitherto.  Besides, 
there  are  several  other  things,  which  have  nothing  of 
Euclid's  accuracy,  and  which  plainly  show,  that  his 
Elements  have  been  much  corrupted  by  unskilful  geo- 
meters; and,  though  these  are  not  so  gross  as  the 
others  now  mentioned,  they  ought  by  no  means  to  re- 
main uncorrected. 

Upon  these  accounts  it  appeared  necessary,  and  I 
hope  will  prove  acceptable,  to  all  lovers  of  accurate 
reasoning,  and  of  mathematical  learning,  to  remove 
such  blemishes,  and  restore  the  principal  Books  of  the 
Elements  to  their  original  accuracy,  as  far  as  I  was 
able;  especially  since  these  Elements  are  the  founda- 
tion of  a  science  by  which  the  investigation  and  dis- 


VI  PREFACE. 


coveiy  of  useful  truths,  at  least  in  mathematical  learn- 
ing, is  promoted  as  far  as  the  limited  powers  of  the 
mind  allow;  and  which  likewise  is  of  the  greatest  use 
in  the  arts  both  of  peace  and  war,  to  many  of  which 
geometry  is  absolutely  necessary.  This  I  have  endea- 
voured to  do,  by  taking  away  the  inaccurate  and  false 
reasonings  which  unskilful  editors  have  put  into  the 
place  of  some  of  the  genuine  Demonstrations  of  Euclid, 
who  has  ever  been  justly  celebrated  as  the  most  accu- 
rate of  geometers,  and  by  restoring  to  him  those  things 
which  Theon  or  others  have  suppressed,  and  which 
have,  these  many  ages,  been  buried  in  oblivion. 

In  this  edition,  Ptolemy's  Proposition  concerning  a 
property  of  quadrilateral  figures  in  a  circle,  is  added  at 
the  end  of  the  sixth  Book.  Also  the  Note  on  the  29th 
Proposition,  Book  1st,  is  altered,  and  made  more  ex- 
plicit, and  a  more  general  Demonstration  is  given,  in- 
stead of  that  which  was  in  the  Note  on  the  10th  Defi- 
nition of  Book  11th;  besides,  the  Translation  is  much 
amended  by  the  friendly  assistance  of  a  learned  gen- 
tleman. 

To  which  are  also  added,  the  Elements  of  Plane  and 
Spherical  Trigonometry,  which  are  commonly  taught 
after  the  Elements  of  Euclid. 


THE 

BfcEMEXTS  OF  EUCLID. 


BOOK  I. 

DEFIiriTIONS. 

I. 

A  POINT  is  that  which  hath  no  parts,  or  which  hath  no  magni- 
tude. :% 

II. 

A  line  is  length  without  breadth.  ^ 

ip.  ^ 

The  extremities  of  a  line  are  points. 

IV. 

A  straight  line  is  that  which  lies  evenly  between  its  extreme 
points.  ^"^ 

V.  * 

A  superficies  is  that  which  hath  only  length  and  breadth. 

VI. 
The  extremities  of  a  superficies  are  lines.  .  #     * 

VI!. 

A  plane  superficies  is  that  in  which  any  two  points  being  taken,* 

the  straight  line  between  them  lies  wholly  in  that  superficies. 

Vill. 
"A  plane  angle  is  the  inclination  of  two  lines  to  one  another*  in 
a  plane,  which  meet  together,  but  are  not  in  the  same  direc- 
tion." 

IX. 
A  plane  rectilineal  angle  is  the  inclination  of  two  straight  lines 
to  one  another,  which  meet  together,  but  are  not  in  the  same 
straight  line, 

A         D 


/ 


B  C  1£ 

'^  See  Notes. 


8 


THE  ELEMENTS  OF  EUCLID. 


BOOK  I. 


N.  B.  'When  several  angles  are  at  one  point  B,  any  one  of  them 
'  is  expressed  by  three  letters,  of  which  the  letter  that  is  at  the 
'  vertex  of  the  angle,  that  is,  at  the  point  in  which  the  straight 
'  lines  that  contain  the  angle  meet  one  another,  is  put  between  the 
'  other  two  letters,  and  one  of  these  two  is  somewhere  upon  one 

*  of  those  straight  lines,  and  the  other  upon  the  other  line:  Thus 

*  the  angle  which  is  contained  by  the  straight  lines  AB,  CB  is 
'  named  the  angle  ABC,  or  CBA;  that  which  is  contained  by  AB, 
'  DB  is  named  the  angle  ABD,  or  DBA^  and  that  which  is  con- 
'  tained  by  DB,  CB  is  called  the  angle  DBC,  or  CBD5  but,  if 
'  there  be  only  one  angle  at  a  point,  it  may  be  expressed  by  the 

*  letter  placed  at  that  point;  as  the  angle  at  E.* 

X. 

When  a  straight  line  standing  on  another 
straight  line  makes  the  adjacent  angles 
equal  to  one  another,  each  of  the  angles 
is  called  a  right  angle:  and  the  straight 
line  which  stands  on  the  other  is  called 
a  perpendicular  to  it. 


XI. 

An  obtuse  angle  is  that  which  is  greater  than  a  right  angle. 


XII. 
An  acute  angle  is  that  which  is  less  than  a  right  angle. 

XIII. 
"A  term  or  boundary  is  the  extremity  of  any  thing." 

XIV. 
A  figure  is  that  which  is  enclosed  by  one  or  more  boundaries. 

XV. 

A  circle  is  a  plane  figure  contained  by  one  line,  which  is  called 
the  circumference,  and  is  such  that  all  straight  lines  drawn  from 
a  certain  point  within  the  figure  to  the  circumference,  are  equal 
to  one  another: 


ROOK  I. 


THE  ELEMENTS  OF  EUCLID. 


XVI. 

And  this  point  is  called  the  centre  of  the  circle. 

XVII. 

A  diameter  of  a  circle  is  a  straight  line  drawn  through  the  centre, 
and  terminated  both  ways  by  the  circumference. 

XVIII. 

A  semicircle  is  the  figure  contained  by  a  diameter  and  the  part 
of  the  circumference  cut  off  by  that  diameter. 

XIX. 

"A  segment  of  a  circle  is  the  figure  contained  by  a  straight  line, 
and  the  circumference  it  cuts  off." 

XX. 

Rectilineal  figures  are  those  which  are  contained  by  straight  lines. 

XXI. 

Trilateral  figures,  or  triangles,  by  three  straight  lines. 

XXII. 

Quadrilateral,  by  four  straight  lines. 

XXIII. 

Multilateral  figures,  or  polygons,  by  more  than  four  straight  lines. 

XXIV. 

Of  three  sided  figures,  an  equilateral  triangle  is  that  which  has 
three  equal  sides. 

XXV. 
An  isosceles  triangle,  is  that  which  has  only  two  sides  equal. 


XXVI. 

A  scalene  triangle,  is  that  which  has  three  unequal  sides. 

XXVII. 

A  right-angled  triangle,  is  that  which  has  a  right  angle. 

XXVIII. 

An  obtuse-angled  triangle,  is  that  which  has  an  obtuse  angle. 


/ 


/ 


/ 


XXIX. 

An  acute-angled  triangle,  is  that  which  has  three  acute  angles. 

XXX. 

Of  four-sided  figures,  a  square  is  that  which  has  all  its  sides  equal, 
and  all  its  angles  right  angles. 
B 


10 


THE  ELEMENTS  01?   EUCLID. 


BOOK  I. 


XXXI. 
An  oblong,  is  that  which  has  all  its  angles  right  angles,  but  has  not 
all  its  sides  equal. 

XXXII. 
A  rhombus,  is  that  which  has  all  its  sides  equal,  but  its  angles 
are  not  right  angles. 


/ 


7 


/ 


/ 


/ 


/ 


/■ 


y 


XXXIII. 

A  rhomboid,  is  that  which  has  its  opposite  sides  equal  to  one 
another,  but  all  its  sides  are  not  equal,  nor  its  angles  right  an- 
gles. 

XXXIV. 

All  other  four  sided  figures  besides  these,  are  called  trapeziums. 

XXXV. 

Parallel  straight  lines,  are  such  as  are  in  the  same  plane,  and 
which  being  produced  ever  so  far  both  ways,  do  not  meet. 


POSTULATES. 
I. 

Let  it  be  granted  that  a  straight  line  may  be  drawn  from  any  one 
point  to  any  other  point. 

II. 
That  a  terminated  straight  line  may  be  produced  to  any  length  in 
a  straight  line. 

III. 
And  that  a  circle  may  be  described  from  any  centre,  at  any  dis- 
tance from  that  centre. 


AXIOMS. 
I. 

Things  which  are  equal  to  the  same  are  equal  to  one  another. 

II. 

If  equals  be  added  to  equals,  the  wholes  are  equal. 


BOOK  I.  THE  ELEMENTS  OF  EUCLID.  1  1 

III. 

If  equals  be  taken  from  equals,  the  remainders  are  equal. 

IV. 

If  equals  be  added  to  unequals,  the  wholes  are  unequal. 

V. 

If  equals  be  taken  from  unequals,  the  remainders  are  unequal. 

VI. 

Things  which  are  double  of  the  same,  are  equal  to  one  another. 

VII. 

Things  which  are  halves  of  the  same,  are  equal  to  one  another. 

VIII. 

Magnitudes  which  coincide  with  one  another,  that  is,  which  ex- 
actly fill  the  same  space,  are  equal  to  one  another. 

IX. 
The  whole  is  greater  than  its  part. 

X. 

Two  straight  lines  cannot  enclose  a  space. 

XI. 

All  right  angles  are  equal  to  one  another. 

XII. 

*'  If  a  straight  line  meet  two  straight  lines,  so  as  to  make  the  two 
"  interior  angles  on  the  same  side  of  it  taken  together  less  than 
"  two  right  angles,  these  straight  lines  being  continually  pro- 
"  duced,  shall  at  length  meet  upon  that  side  on  which  are  the 
"  angles  which  are  less  than  two  right  angles.  See  the  notes 
"  on  prop.  29.  of  Book  I.'* 


12 


THE  ELEMENTS  OF  EUCLID. 


BOOK  I. 


PROPOSITION  I.    PROBLEM. 

To  describe  an  equilateral  triangle  upon  a  given 
finite  straight  line. 

Let  AB  be  the  given  straight  line  5  it  is  required  to  describe 
an  equilateral  triangle  upon  it. 

From  the  centre  A,  at  the  distance 
AB,  describe  (3.  Postulate.)  the  cir- 
cle BCD,  and  from  the  centre  B,  at 
the  distance  BA,  describe  the  circle 
ACE;  and  from  the  point  C,  in 
which  the  circles  cut  one  another, 
draw  the  straight  lines  (2.  Post.) 
CA,  CB  to  the  points  A,  B;  ABC 
shall  be  an  equilateral  triangle. 

Because  the  point  A  is  the  centre  of  the 
equal  (15.  Definition.)  to  AB;  and  because  the  point  B  is  the  cen 
tre  of  the  circle  ACE,  BC  is  equal  to  BA:  but  it  has  been  proved 
that  CA  is  equal  to  AB;  therefore  CA,  CB  are  each  of  them 
equal  to  AB;  but  things  which  are  equal  to  the  same  are  ecjual 
to  one  another;  (1st  Axiom.)  therefore  CA  is  equal  to  CB; 
wherefore  CA,  AB,  BC  are  equal  to  one  another;  and  the  triangle 
ABC  is  therefore  equilateral,  and  it  is  described  upon  the  given 


circle  BCD,  AC  is 


straight  line  AB. 


Which  was  required  to  be  done. 


PROP.  II.    PROB. 

From  a  given  point  to  draw  a  straight  line  equal  to 
a  given  straight  line. 

Let  A  be  the  given  point,  and  BC  the  given  straight  line;  it  is 
required  to  draw  from  the  point  A  a  straight  line  equal  to  BC. 

From  the  point  A  to  B  draw  (1. 
Post.)  the  straight  line  AB;  and 
upon  it  describe  (I.  1.)  the  equilate- 
ral triangle  DAB,  and  produce  (2. 
Post.)  the  straight  lines  DA,  DB, 
to  E  and  F;  from  the  centre  B,  at 
the  distance  BC,  describe  (3.  Post.) 
the  circle  CGFI,  and  from  the  cen- 
tre D,  at  the  distance  DG,  describe 
the  circle  GKL.  AL  shall  be  equal 
to  BC. 

Because  the  point  B  is  the  centre 
of  the  circle  CGH,  BC  is  equal  (15. 
Def.)  to  BG;  and  because  D  is  the  centre  of  the  circle  GKL,  DL 
is  equal  to  DG,  and  DA,  DB,  parts  of  them,  are  equal:  therefore 
the  remainder  AL  is  equal  to  the  remainder  (3.  Ax.)  BG;  but  it 
has  been  shown,  that  BC  is  equal  to  G,  wherefore  AL  and  BC 
are  each  of  them  equal  to  BG;  ajid  things  that  are  equal  to  the 


BOOK  I. 


THE  ELEMENTS  OF  EUCLID. 


13 


same  are  equal  to  one  another;  therefore  the  straight  line  AL  is 
equal  to  BC.  Wherefore  from  the  given  point  A  a  straight  line 
AL  has  been  drawn  equal  to  the  given  straight  line  BC.  Which 
was  to  be  done. 

> 

PROP  III.    PROB. 

From  the  greater  of  two  given  straight  Hues  to  cut 
off  a  part  equal  to  the  less. 

Let  AB  and  C  be  the  two  given 
straight  lines,  whereof  AB  is  the 
greater.  It  is  required  to  cut  off 
from  AB,  the  greater,  a  part  equal 
to  C,  the  less. 

From  the  point  A  draw  (2.  1.)  the 
straight  line  AD  equal  to  C;  and 
from  the  centre  A,  and  at  the  dis- 
tance AD,  describe  (3.  Post.)  the 
circle  DEF;  and  because  A  is  the 
centre  of  the  circle  DEF,  AE  shall  be  equal  to  AD;  but  the 
straight  line  C  is  likewise  equal  to  AD;  whence  AE  and  C  are 
each  of  them  equal  to  AD:  wherefore  the  straight  line  AE  is 
equal  to  (1.  Ax.)  C,  and  from  AB,  the  greater  of  two  straight 
lines,  a  part  AE  has  been  cut  off  equal  to  C  the  less.  Which  was 
to  be  done.  §; 

PROP.  IV.  THEOREM. 

If  two  triangles  have  two  sides  of  the  one  equal  to 
two  sides  of  the  other,  each  to  each;  and  have  like- 
wise the  angles  contained  by  those  sides  equal  to  one 
another,  they, shall  likewise  have  their  bases,  or  third 
sides,  equal;  and  the  two  triangles  shall  be  equal;  and 
their  other  angles  shall  be  equal,  each  to  each,  viz. 
those  to  which  the  equal  sides  are  opposite. 

Let  ABC,  DEF  be  two  triangles  which  have  the  two  sides  AB 
AC  equal  to  the  two  sides  DE,  DF,  each  to  each,  viz.  AB  to  DE, 
and  AC  to  DF;  and  the  angle     A  D 

BAC  equal  to  the  angle  EDF, 
the  base  BC  shall  be  equal  to 
the  base  EF;  and  the  triangle 
ABC  to  the  triangle  DEF;  and    ; 
the  other  angles,  to  which  the    / 
equal  sides  are  opposite,  shall  ' 
be  equal  each  to  each,  viz.  the 
angle  ABC  to  the  angle  DEF, ' 
and  the  angle  ACB  to  DFE.     '  B  C     E  F 

For,  if  the  triangle  ABC  be  applied  to  DEF,  so.  that  the  point 
A  may  be  on  D,  and  the  straight  line  AB  upon  DE;  the  point  B 
shall  coincide  with  the  point  E,  because  AB  is  equal  to  DE;  and 
AB  coinciding  with  DE,  AC  shall  coincide  with  DF;  because 


14  THE  ELEMENTS  OF    EUCLID.  BOOK  I. 

the  angle  BAG  is  equal  to  the  angle  EDF;  wherefore  also  the 
point  C  shall  coincide  with  the  point  F,  because  the  straight  line 
AC  is  equal  to  DF:  but  the  point  B  coincides  with  the  point  E; 
wherefore  the  base  BC  shall  coincide  with  the  base  EF,  because 
the  point  B  coinciding  with  E,  and  C  with  F,  if  the  base  BC  does 
not  coincide  with  the  base  EF,  two  straight  lines  would  inclose  a 
space,  which  is  impossible.  (10.  Ax.)  Therefore  the  base  BC 
shall  coincide  with  the  base  EF,  and  be  equal  to  it.  Wherefore 
the  whole  triangle  ABC  shall  coincide  with  the  whole  triangle 
DEF,  and  be  equal  to  it;  and  the  other  angles  of  the  one  shall  co- 
incide with  the  remaining  angles  of  the  other,  and  be  equal  to 
them,  viz.  the  angle  ABC  to  the  angle  DEF,  and  the  angle  ACB 
to  DFE.  Therefore,  if  two  triangles  have  two  sides  of  the  one 
equal  to  two  sides  of  the  other,  each  to -each,  and  have  likewise 
the  angles  contained  by  those  sides  equal  to  one  another,  their 
bases  shall  likewise  be  equal,  and  the  triangles  be  equal,  and  their 
other  angles  to  which  the  equal  sides  are  opposite  shall  be  equal, 
each  to  each.     Which  was  to  be  demonstrated. 

PROP.  V.  THEOR. 

The  angles  at  the  base  of  an  isosceles  triangle  are 
equal  to  one  another:  and,  if  the  equal  sides  be  pro- 
duced, the  angles  upon  the  other  side  of  the  base  shall 
be  equal. 

Let  ABC  be  an  isosceles  triangle,  of  which  the  side  AB  is 
equal  to  AC,  and  let  the  straight  lines  AB,  AC  be  produced  to 
D  and  E,  the  angle  ABC  shall  be  equal  to  the  angle  ACB,  and 
the  angle  CBD  to  the  angle  BCE. 

In  BD  take  any  point  F,  and  from  AE  the  greater,  cut  off 
AG  equal  (3.  1.)  to  AF,  the  less,  and  join  FC,  GB. 

Because  AF  is  equal  to  AG,  and  AB  to  AC,  the  two  sides  FA, 
AC,  are  equal  to  the  two  GA,  AB,  each  to  each;  and  they  con 
tain  the  angle  FAG  common  to  the  two 
triangles  AFC,  AGB;  therefore  the  base 
FC  is  equal  (4.  1.)  to  the  base  GB,  and  the 
triangle  AFC  to  the  triangle  AGB;  and 
the  remaining  angles  of  the  one  are  equal 
(4.  1.)  to  the  remaining  angles  of  the 
other,  each  to  each,  to  which  the  equal 
sides  are  opposite;  viz.  the  angle  ACF  to 
the  angle  ABG,  and  the  angle  AFC  to  the 
angle  AGB;  and  because  the  whole  AF  is 
equal  to  the  whole  AG,  of  which  the  parts 
AB,  AC,  are  equal:  the  remainder  BF 
shall  be  equal  (S.  Ax.)  to  the  remainder  j^ 
CG;  and  FC  was  proved  to  be  equal  to 
GB;  therefore  the  two  sides  BF,  FC  are  equal  to  the  two  CG, 
GB,  each  to  each:  and  the  angle  BFC  is  equal  to  the  angle  CGB, 
and  the  base  BC  is  common  to  the  two  triangles  BFC,  CGB; 
wherefore  the  triangles  are  equal  (4.  I.)  and  their  remaining  an- 


BOOK  I.  THE  ELEMENTS  OF  EUCLID.  15 

gles,  each  to  each,  to  which  the  equal  sides,  are  opposite;  there- 
fore the  angle  FBC  is  equal  to  the  angle  GCB,  and  the  angle  BCF 
to  the  angle  CBG;  and,  since  it  has  been  demonstrated,  that  the 
whole  angle  ABG  is  equal  to  the  whole  ACF,  the  parts  of  which, 
the  angles  CBG,  BCF  are  also  equal;  the  remaining  angle  ABC 
is  therefore  equal  to  the  remaining  angle  ACB,  which  are  the 
angles  at  the  base  of  the  triangle  ABC:  and  it  has  also  been  proved 
that  the  angle  FBC  is  equal  to  the  angle  GCB,  which  are  the  an- 
gles upon  the  other  side  of  the  base.  Therefore  the  angles  at  the 
base,  &c.  Q.  E.  D. 

Corollary.  Hence  every  equilateral  triangle  is  also  equian- 
gular. 

PROP.  VI.  THEOR. 

If  two  angles  of  a  triangle  be  equal  to  one  another, 
the  sides  also  which  subtend,  or  are  opposite  to,  the 
equal  angles,  shall  be  equal  to  one  another. 

Let  ABC  be  a  triangle  having  the  angle  ABC  equal  to  the 
angle  ACB;  the  side  AB  is  also  equal  to  the  side  AC. 

For  if  AB  be  not  equal  to  AC,  ont  of  them  is  greater  than  the 
other;  let  AB  be  the  greater,  and  from  it  cut  (3.  1.)  off  DB  equal 
to  AC  the  less,  and  join  DC;  therefore,  because 
in  the  triangles  DBC,  ACB,  DB  is  equal  to  AC, 
and  BC  common  to  both,  the  two  sides  DB,  BC 
are  equal  to  the  two  AC,  CB,  each  to  each;  and 
the  angle  DBC  is  equal  to  the  angle  ACB; 
therefore  the  base  DC  is  equal  to  the  base  AB, 
and  the  triangle  DBC  is  equal  to  the  triangle 
(4.  1.)  ACB,  the  less  to  the  greater;  v.'hich  is 
absurd.  Therefore  AB  is  not  unequal  to  AC, 
that  is,  it  is  equal  to  it.  Wherefore,  if  two  / 
angles,  &c.  Q.  E.  D.  ^ 

B 

Cor.  Hence  every  equiangular  triangle  is  also  equilateral. 

PROP.  Vn.  THEOR. 

Upon  the  same  base,  and  on  the  same  side  of  it, 
there  cannot  be  two  triangles  that  have  their  sides 
which  are  terminated  in  one  extremity  of  the  base  equal 
to  one  another,  and  likewise  those  which  are  terminated 
in  the  other  extremity.* 

If  it  be  possible,  let  there  be  tv/o  triangles  ACB,  ADB,  upon 
the  same  base  AB,  and  upon  the  same  side  of  it,  which  have  their 
sides  CA,  DA,  terminated  in  the  extremity  A  of  the  base  equal 
to  one  another,  and  likewise  their  sides,  CB,  DB,  that  are  termi- 
nated in  B. 

*  See  Note. 


16 


THE  ELEMENTS  OF  EUCLID. 


BOOK  I. 


Join  CD;  then,  in  the  case  in  which  the 
vertex  of  each  of  the  triangles  is  without 
the  other  triangle,  because  AC  is  equal  to 
AD,  the  angle  ACU  is  equal  (5.  1.)  to  the 
angle  ADC:  but  the  angle  ACD  is  greater 
than  the  angle  BCD;  therefore  the  angle 
ADC  is  gr'^ater  also  than  BCD;  much 
more  then  is  the  angle  BDC  greater  than 
the  angle  BCD.  Again,  because  CB  is 
equal  to  DB,  the  angle  BDC  is  equal  (5.1.)  ^ 


C     D 


B 


to  the  angle  BCD;  but  it  has  been  demon-  ^  ■" 

strated  to  be  greater  than  it;  which  is  impossible. 

But  if  one  of  the  vertices,  as  D,  be  within  the  other  triangle 
ACB;  produce  AC,  AD  to  E,  F;   there-  E 

fore,  because  AC  is  equal  to  AD  in  the 
triangle  ACD,  the  angles  ECD,  FDC  upon 
the  other  side  of  the  base  CD  are  equal 
(5.  1.)  to  one  another,  but  the  angle  ECD 
is  greater  than  the  angle  BCD;  wherefore 
the  angle  FDC  is  likewise  greater  than 
BCD;  much  more  then  is  the  angle  BDC 
greater  than  the  angle  BCD.     Again,  be- 
cause CB  is  equal  to  DB,  the  angle  BDC    /V 
is  equal  (5. 1.)  to  the  angle  BDC;  but  BCD  /' 
has  been  proved  to  be   greater  than  the 
same   BCD;    which   is   impossible.     The  ^ 
case  in  which  the  vertex  of  one  triangle  is  upon  a  side  of  the 
other,  needs  no  demonstration. 

Therefore  upon  the  same  base,  and  on  the  same  side  of  it,  there 
cannot  be  two  triangles  that  have  their  sides  which  are  termi- 
nated in  one  extremity  of  the  base  equal  to  one  another,  and  like- 
wise those  which  are  terminated  in  the  other  extremity.  Q.  E.  D. 

PROP.  VIII.    THEOR. 

If  two  triangles  have  two  sides  of  the  one  equal  to 
two  sides  of  the  other,  each  to  each,  and  have  likewise 
their  bases  equal;  the  angle  which  is  contained  by  the 
two  sides  of  the  one  vshall  be  equal  to  the  angle  con- 
tained by  the  two  sides  equal  to  them,  of  the  other. 

Let  ABC,  DEF  be  A 
two  triangles,  having    ^ 
the  two  sides  AB,  AC,  \ 
equal  to  the  two  sides    \ 
DE,  DF,  each  to  each, 
viz.   AB  to  DE,   and 
AC  to  DF;    and  also 
the  base  BC  equal  to 
the  base  EF.    The  an- 
gle  BAC   is  equal  to 
the  angle  EDF. 


D     G 


BOOK  I. 


THE  ELEMENTS  OF  EUCLID. 


ir 


For,  if  the  triangle  ABC  be  applied  to  DEF,  so  that  the  point 
B  be  on  E,  and  the  straight  line  BC  upon  EF;  the  point  C  shall 
also  coincide  with  the  point  F.  Because  BC  is  equal  to  EFj 
therefore  BC  coinciding  with  EF,  BA  and  AC  shall  coincide  with 
ED  and  DF;  for,  if  the  base  BC  coincides  with  the  base  EF,  but 
the  sides  BA,  CA  do  not  coincide  with  the  sides  ED,  FD,  but 
have  a  different  situation,  as  EG,  FG^  then  upon  the  same  base 
EF,  and  upon  the  same  side  of  it,  there  can  be  two  triangles  that 
have  their  sides  which  are  ternfiinated  in  one  extremity  of  the 
base  equal  to  one  another,  and  likewise  their  sides  terminated  in 
the  other  extremity;  but  this  is  impossible;  (7.  1.)  therefore,  if 
the  base  BC  coincides  with  the  base  EF,  the  sides  BA,  AC  can- 
not but  coincide  with  the  sides,  ED,  DF;  wherefore  likewise  the 
angle  BAC  coincides  with  the  angle  EDF,  and  is  equal  (8.  Ax.) 
to  it.     Therefore,  if  two  triangles,  &c.  Q.  E.  D. 


PROP.  IX.    PROB. 


To  bisect  a  given  rectilineal  angle,  that  is,  to  divide 
it  into  two  equal  angles. 


Let  BAC  be  the  given  rectilineal  angle,  it  is  required 
sect  it. 

Take  any  point  D  in  AB,  and  from  AC  cut  (3.  I.)  off  AE 
to  AD;  join  DE,  and  upon  it  describe  (l.  I.) 
an  equilateral  triangle  DEF;  then  join  AF; 
the  straight  line  AF  bisects  the  angle  BAC. 

Because  AD  is  equal  to  AE,  and  AF  is 
common  to  the  two  triangles  DAF,  EAF; 
the  two  sides  DA,  AF,  are  equal  to  the  two 
sides,  EA,  AF,  each  to  each;  and  the  base 
DF  is  equal  to  the  base  EF;  therefore  the 
angle  DAF  is  equal  (8.  1.)  to  the  angle 
EAF;  wherefore  the  given  rectilineal  angle 
BAC  is  bisected  by  the  straight  line  AF, 
which  was  to  be  done. 


to  bi- 
equal 


PROP.  X.    PROB. 

To  bisect  a  given  finite  straight  line,  that  is,  to  di- 
vide it  into  two  equal  parts. 

Let  AB  be  the  given  straight  line:  it  is  required  to  divide  it 
into  two  equal  parts. 

Describe  (l.  I.)  upon  it  an  equilateral  triangle  ABC,  and  bisect 
(9.  1.)  the  angle  ACB  by  the  straight  line  CD.     AB  is  cut  into 
two  equal  parts  in  the  point  D. 
C 


18 


THE  ELEMENTS  OF  EUCLID. 


BOOK  I. 


Because  AC  is  equal  to  CB,  and  CD  com- 
mon to  the  two  triangles  ACD,  BCDj  the 
two  sides  AC,  CD  are  equal  to  BC,  CD,  each 
to  each;  and^he  angle  ACD  is  equal  to  the 
angle  BCD;  th'erefore  the  base  AD  is  equal 
to  the  base  (4.  1.)  DB  and  the  straight  line 
AB  is  divided  into  two  equal  parts  in  the 
point  D.     Which  was  to  be  done. 


C 


PROP.  XI.    PROB. 

To  draw  a  straight  line  at  right  angles  to  a  given 
straight  line,  from  a  given  point  in  the  same. 

Let  AB  be  a  given  straight  line,  and  C  a  point  given  in  it:  it  is 
required  to  draw  a  straight  line  from  the  point  C  at  right  angles 
to  AB.*  ' 

Take  any  point  D  in  AC,  and  (3.  1.)  make  CE  equal  lo  CD, 
and  upon  DE  describe  (l.  1.)  the  equi-  F 

lateral  triangle,  DFE,  and  join  FC; 
the  straight  line  FC  drawn  from  the 
given  point  C  is  at  right  angles  to  the 
given  straight  line  AB. 

Because  DC  is  equal  to  CE,  and  FC 
common  to  the  two  triangles  DCF, 
ECF;  the  two  sides  DC,  CF,  are  equal 
to  the  two  EC,  CF,  each  to  each;  and  ^ 
the  base  DF  is  equal  to  the  base  EF;  therefore  the  angle  DCF  is 
equal  (8.  1.)  to  the  angle  ECF;  and  they  are  adjacent  angles. 
But,  when  the  adjacent  angles  which  one  straight  line  makes  with 
another  straight  line  are  equal  to  one  another,  each  of  them  is 
called  a  right  (10.  Def.  1.)  angle;  therefore  each  of  the  angles 
DCF,  ECF,  is  a  right  angle.  Wherefore,  from  the  given  point 
C,  in  the  given  straight  line  AB,  FC  has  been  drawn  at  right  an- 
gles to  AB.     Which  was  to  be  done. 

Cor.  By  help  of  this  problem,  it  may  be  demonstrated,  that  two 
straight  lines  cannot  have  a  common  segment. 

If  it  be  possible,  let  the  two  straight  lines  ABC,  ABD  have  the 
segment  AB  common  to  both  of  them.  From  the  point  B  draw 
BE  at  right  angles  to  AB;  and  because  ABC  is  a  straight  line, 
the  angle  CBE  is  equal  (10.  Def.  E 

1.)  to  the  angle  EBA;  in  the 
same  manner,  because  ABD  is  a 
straight  line,  the  angle  DBE  is 
equal  to  the  angle  EiBA;  where- 
fore the  angle  DBE  is  equal  to 
the  angle  CBE,  the  less  to  the 
greater;  which  is  impossible;  — 
therefore  two  straight  lines  can- ,^ 
not  have  a  common  segment. 


B 


See  Note. 


BOOK  I. 


THE  ELEMENTS  OE  EUCLID. 


19 


PROP.  XII.  PROB. 

To  draw  a  straight  line  perpendicular  to  a  given 
straight  line  of  an  unlimited  length,  from  a  given  point 
without  it. 

Let  AB  be  the  given  straight  line,  which  may  be  produced  to 
any  length  both  ways,  and  let  C  be  a  point  without  it.  It  is  re- 
quired to  draw  a  straight  line  per-  C 
pendicular  to  AB  from  the  point  C, 

Take  any  point  D  upon  the  other     i 
side  of  AB,  and  from  the  centre  C,     \ 
at   the    distance,   CD,  describe   (3. 
Post.)  the  circle  FDG  meeting  AB 
in  F,  G,-  and,  bisect  (lO.  1.)  FG  in 


H,  and  join  CF,  CH,  CGj  the 
straierht  line  CH,  drawn  from  the 
given  point  C,  is  perpendicular  to  the  given  straight  line  AB. 

Because  FH  is  equal  to  HG,  and  HC  common  to  the  two  tri- 
angles FHC,  GHC,  the  two  sides  FH,  HC  are  equal  to  the  two 
GH,  HC,  each  to  eachj  and  the  base  CF  is  equal  (15.  Def.  1.)  to 
the  base  CG,-  therefore  the  angle  CHF  is  equal  (8.  1.)  to  the  an- 
gle CHG;  and  they  are  adjacent  angles;  but  when  a  straight  line 
standing  on  a  straight  line  makes  the  adjacent  angles  equal  to  one 
another,  each  of  them  is  a  right  angle,  and  the  straight  line  which 
stands  upon  the  other  is  called  a  perpendicular  to  it;  therefore 
from  the  given  point  C  a  perpendicular  CH  has  been  drawn  to 
the  given  straight  line  AB.     Which  was  to  be  done. 

PROP.  XIII.  THEOR. 

The  angles  which  one  straight  line  makes  with  an- 
other upon  the  one  side  of  it,  are  either  two  right  an- 
gles, or  are  together  equal  to  two  right  angles. 

Let  the  straight  line  AB  make  with  CD,  upon  one  side  of  it, 
the  angles  CBA,  ABD;  these  are  either  two  right  angles,  or  are 
together  equal  to  two  right  angles. 

For,  if  the  an8:le  CBA  be  equal  to  ABD,  each  of  them  is  a  right 
A  E  A 


D 


B 


C 


D       B 


(def.  10.)  angle;  but,  if  not,  from  the  point  B  draw  BE  at  right 
angles  (11.  h)  to  CD;  therefore  the  angles  CBE,  EBD,  are  two 
right  angles;  (def.  10.)  and  because  CBE  is  equal  to  the  two  an- 


20  THE  ELEMENTS  OF  EUCLID.  BOOK  I. 

gles  CBA,  ABE  together,  add  the  angle  EBD  to  each  of  these 
equals;  therefore  the  angles  CBE,  EBD  are  (2.  Ax.)  equal  to  the 
three  angles  CBA,  ABE,  EBD.  Again,  because  the  angle  DBA 
is  equal  to  the  two  angles  DBE,  EBA,  add  to  these  equals  the  an- 
gle ABC;  therefore  the  angles  DBA,  ABC  are  equal  to  the  three 
angles  DBE,  EBA,  ABC;  but  the  angles  CBE,  EBD  have  been 
demonstrated  to  be  equal  to  the  same  three  angles;  and  things 
that  are  equal  to  the  same  are  equal  (l.  Ax.)  to  one  another; 
therefore  the  angles  CBE,  EBD  are  equal  to  the  angles  DBA, 
ABC:  but  CBE  EBD  are  two  right  angles:  therefore  DBA,  ABC 
are  together  equal  to  two  right  angles.  Wherefore,  when  a 
straight  line,  &,c.  Q.  E.  D. 

PROP.  XIV.  THEOR. 

If,  at  a  point  in  a  straight  line,  two  other  straight 
Hnes,  upon  the  opposite  sides  of  it,  make  the  adjacent 
angles  together  equal  to  two  right  angles,  these  two 
straight  lines  shall  be  in  one  and  the  same  straight  line. 


/ 


At  tho  point  B  in  the  straight 
line  AB,  let  the  two  straight  lines 
BC,  BD  upon  the  opposite  sides 
of  AB  make  the  adjacent  angles 
ABC,  ABD  equal  together  to  two 
right  angles,  BD  is  in  the  same 
straight  line  with  CB. 

For,  if  BD  be  not  in  the  same 
straight  line  with  CB,  let  BE  be  ^ 
in  the  same  straight  line  with  it;  therefore,  because  the  straight 
line  AB  makes  angles  with  the  straight  line  CBE,  upon  one  side 
of  it,  the  angles  ABC,  ABE  are  together  equal  (13.  1.)  to  two 
right  angles;  but  the  angles  ABC,  ABD  are  likewise  together 
equal  to  two  right  angles;  therefore  the  angles  CBA,  ABE  are 
equal  to  the  angles  CBA,  ABD:  take  away  the  common  angle 
ABC,  the  remaining  angle  ABE  is  equal  (3.  Ax.)  to  the  remain- 
ing angle  ABD,  the  less  to  the  greater,  which  is  impossible;  there- 
fore BE  is  not  in  the  same  straight  line  with  BC.  And,  in  like 
manner,  it  may  be  demonstrated  that  no  other  can  be  in  the  same 
straight  line  with  it  but  BD,  which  therefore  is  in  the  same 
straight  line  with  CB.     Wherefore,  if  at  a  point,  8cc.  Q.  E.  D. 

PROP.  XV.  THEOR. 

If  two  straight  lines  cut  one  another,  the  vertical  or 
opposite  angles  shall  be  equal. 

Let  the  two  straight  lines  AB,  CD  cut  one  another  in  the  point 
E;  the  angle  AEC  shall  be  equal  to  the  angle  DEB,  and  CEB  to 
AED. 


BOOK  I. 


THE  ELEMENTS  OF  EUCLID. 


21 


Because  the  straight  line  AE  makes 
with  CD  the  angles  CEA,  AED,  these  C 
angles  are  together  equal  (13.    1.)  to 
two  right  angles.     Again,  because  the 


straight  line  DE  makes  with  AB  the  A 
angles  AED,  DEB,  these  also  are  to- 
gether equal  (13.  1.)  to  two  right  an- 
gles; and  CEA,  AED  have  been  de-  D 
monstrated  to  be  equal  to  two  right  angles;  wherefore  the  angles 
CEA,  AED  are  equal  to  ihe  angles  AED,  DEB.  Take  away  the 
common  angle  AED,  and  the  remaining  angle  CEA  is  equal  (3. 
Ax.)  to  the  remaining  angle  DEB.  In  the  same  manner  it  can  be 
demonstrated  that  the  angles  CEB,  AED  are  equal.  Therefore, 
if  two  straight  lines,  Sec.  Q.  E.  D. 

Cor.  1.  From  this  it  is  manifest,  that  if  two  straight  lines  cut 
one  another,  the  angles  they  make  at  the  point  where  they  cut, 
are  together  equal  to  four  right  angles. 

Cor.  2.  And  consequently  that  all  the  angles  made  by  any  num- 
ber of  lines  meeting  in  one  point,  are  together  equal  to  four  right 
angles. 


PROP.  XVI.  THEOR. 


If"  one  side  of  a  triangle  be  produced,  the  exterior 
angle  is  greater  than  either  of  the  interior  opposite  an- 
gles. 

Let  ABC  be  a  triangle,  and  let  its  side  BC  be  produced  to  D, 
the  exterior  angle  ACD  is  greater  than  either  of  the  interior  op- 
posite angles  CBA,  BAC.  A 

Bisect  (10.  1.)  AC  in  E,  join  BE 
and  produce  it  to  F,  and  make  EF 
equal  to  BE;  join  also  FC,  and  pro- 
duce AC  to^. 

Because  AE  is  equal  to  EC,  and 
BE  to  EF;  AE,  EB  are  equal  to 
CE,  EF,  each  to  each;  and  the  an- 
gle AEB  is  equal  (15.  1.)  to  the  an-  i-. 
gle  CEF,  because  they  are  opposite  B 
vertical  angles;  therefore  the  base 
AB  is  equal  (4.  I.)  to  the  base  CF, 
and  the  triangle  AEB  to  the  trian- 
gle CEF,  and  the  remaining  angles, 
to  the  remaining  angles,  each  to 
each,  to  which  the  equal  sides  are  opposite;  wherefore  the  angle 
BAE  is  equal  to  the  angle  ECF;  but  the  angle  ECD  is  greater 
than  the  angle  ECF;  therefore  the  angle  ACD  is  greater  than 
BAE:  in  the  same  manner,  if  the  side  BC  be  bisected,  it  may  be 
demonstrated  that  the  angle  BCG,  that  is,  (15.  1.)  the  angle  ACD, 
is  greater  than  the  angle  ABC.  Therefore,  if  one  side,  Sec.  Q. 
E.  D. 


22      '  THE  ELEMENTS  OF  EUCLID.  BOOK  I. 


PROP.  XVII.  THEOR. 

Any  two  angles  of  a  triangle  are  together  less  than 
two  right  angles. 

Let  ABC  be  any  triangle;  any  two  of 
its  angles  together  are  less  than  two  right 
angles. 

Produce  BC  to  D;  and  because  ACD 
is  the  exterior  angle  of  the  triangle  ABC, 
ACD  is  greater  (16.  I.)  than  the  interior 
and  opposite  angle  ABC;  to  each  of  these 
add  the  angle  ACB;  therefore  the  angles 
ACD,  ACB  are  greater  than  the  angles  B  CD 

ABC,  ACB;  but  ACD,  ACB  are  together  equal  (13.  1.)  to  two 
right  angles;  therefore  the  angles  ABC,  BCA  are  less  than  two 
right  angles.  In  like  manner,  it  may  be  demonstrated,  that  BAC, 
ACB,  as  also  CAB,  ABC,  are  less  than  two  right  angles.  There- 
fore any  two  angles,  Sec.  Q.  E.  D. 


PROP.  XVIII.  THEOR. 

The  greater  side  of  every  triangle  is  opposite  to  the 
greater  angle. 

Let  ABC  be  a  triangle,  of  which  the  A 

side  AC  is  greater  than  the  side  AB; 
the  angle  ABC,  is  also  greater  than 
the  angle  BCA. 

Because  AC  is  greater  than  AB, 
make  (3.  I.)  AD  equal  to  AB,  and 
join  BD;  and  because  ADB  is  the  ex- 
terior angle  of  the  triangle  BDC,  it  is 
greater  (16.  1.)  than  the  interior  and  opposite  angle  DCB;  but 
ADB  is  equal  (5.  1.)  to  ABD,  because  the  side  AB  is  equal  to  the 
side  AD ;  therefore  the  angle  ABD  is  likewise  greater  than  the 
angle  ACB;  wherefore  much  more  is  the  angle  ABC  greater 
than  ACB.     Therefore  the  greater  side.  Sec.  Q.  E.  D. 


PROP.  XIX.  THEOR. 

The  greater  angle  of  every  triangle  is  subtended  by 
the  greater  side,  or  has  the  greater  side  opposite  to  it* 

> 

Let  ABC  be  a  triangle,  of  which  the  angle  ABC  is  greater 
than  the  angle  BCA;  the  side  AC  is  likewise  greater  than  the 
side  AB. 


BOOK  I.  THE  E£>4J!MENTS.0F  EUCLID. 

For,  if  it.'^  not  greater,  AC  must  ■;  A 

either  be  eqiiaJ  to  AB,  or  less  than  it; 
it  is  not  equal,  because  then  the  angle 
ABC  would  be  equal  (5.  1.)  to  the  an- 
gle ACB;  but  it'fsnot;  therefore  AC 
is  not  equal  to  i\B;  neither  is. it  less; 
because  then  the  angle  ABC  would  be 
less  (18.  1.)  than  the  angle  ACB;  but  B '•  C 

it  is  not;  therefore  the  side  AC  is  not  less  than  AB;  and  it  has 
been  shown  that  it  is  not  equal  ot  AB;  therefore  AC  is  greater 
than  AB..    Wherefore  the  greater  angle,  8cc.  Q.  E.  D. 

PROP.  XX.  THEOR. 

Any  two  sides  of  a  triangle  are  together  greater  than 
the  third  side.* 

Let  ABC  be  a  triangle;  any  two  sides  of  it  together  are  greater 
than  the  third  side,  viz.  the  sides  BA,  AC  greater  than  the  side 
BC;  and  AB,  BC  greater  than  AC;  and  BC,  CA  greater  than 
AB. 

Produce    BA  to    the  point  D,  and  D 

make  (3.  1.)  AD  equal  to  AC;  and 
join  DC.  A 

Because  DA  is  equal  to  AC,  the  an- 
gle ADC  is  likewise  equal  (5.  1.)  to 
ACD;  but  the  angle  BCD  is  greater 
than  the  angle  ACD;  therefore  the  an- 
gle BCD  is  greater  than  the  angle  ^ 
ADC;  and  because  the  angle  BCD  of  the  triangle  DCB  is  greater 
than  its  angle  BDC,  and  that  the  greater  (19.  1.)  side  is  opposite 
to  the  greater  angle:  therefore  the  side  DB  is  greater  than  the 
side  BC;  but  DB  is  equal  to  BA  and  AC;  therefore  the  sides 
BA,  AC  are  greater  than  BC.  In  the  same  manner  it  may  be 
demonstrated,  that  the  sides  AB,  BC  are  greater  than  CA,  and 
BC,  CA  greater  than  AB.  Therefore  any  two  sides.  Sec.  Q. 
E.  D. 

PROP.  XXI.  THEOR. 

If,  from  the  ends  of  the  side  of  a  triangle,  there  be 
drawn  two  straight  lines  to  a  point  within  the  triangle, 
these  shall  be  less  than  the  other  two  sides  of  the  tri- 
angle, but  shall  contain  a  greater  angle."^ 

Let  the  two  straight  lines  BD,  CD  be  drawn  from  B,  C,  the 
ends  of  the  side  BC  of  the  triangle  ABC,  to  the  point  D  within 
it;  BD  and  DC  are  less  than  the  other  two  sides  BA,  AC  of  the 
triangle,  but  contain  an  angle  BDC  greater  than  the  angle  BAC. 

Produce  BD  to  E;  and  because  two  sides  of  a  triangle  are 
greater  than  the  third  side,  the  two  sides  BA,  AE  of  the  triangle 

"  See  Notes. 


24 


THE  ELEMENTS  OF  EUCLID. 


BOOK  I. 


ABE  are  greater  than  BE.     To  each  of  these  add  EC5  therefore 
the  sides   BA,  AC   are  greater  than 


A 


BE,  EC: 
CE,   ED 


again, 
of   the 


because  the  two  sides 
triangle   CED    are 


greater  than  CD,  add  DB  to  each  of 
these;  therefore  the  sides  CE,  EB  are 
greater  than  CD,  DB;  but  it  has  been 
shown  that  BA,  AC  are  greater  than 
BE,  EC;  much  more  then  are  BA, 
AC  greater  than  BD,  DC. 

Again,  because  the  exterior  angle 
of  a  triangle  is  greater  than  the  interior  and  opposite  angle,  the 
exterior  angle  BDC  of  the  triangle  CDE  is  greater  than  CED; 
for  the  same  reason,  the  exterior  angle  CEB  of  the  triangle  ABE 
is  greater  than  BAC;  and  it  has  been  demonstrated  that  the  angle 
BDC  is  greater  than  the  angle  CEB;  much  more  then  is  the  an- 
gle BDC  greater  than  the  angle  BAC.  Therefore,  if  from  the 
ends  of,  he.  Q.  E.  D. 


PROP.  XXII.    PROB. 

To  make  a  triangle  of  which  the  sides  shall  be  equal 

to  three  given  straight  lines,  but  any  two  whatever  of 

these  must  be  greater  than  the  third  (20.  1.)^ 

Let  A,  B,  C  be  the  three  given  straight  lines,  of  which  any  two 
whatever  are  greater  than  the  third,  viz.  A  and  B  greater  than  C; 
A  and  C  greater  than  B;  and  B  and  C  than  A.  It  is  required  to 
make  a  triangle  of  which  the  sides  shall  be  equal  to  A,  B,  C,  each 

to  each. 

Take  a  straight  line  DE  terminated  at  Ihe  ppint  D,  but  unli- 

/' 


mited  towards  E,  and  make 
(3.  1.)  DF  equal  to  A,  FG  to 
B,  and  GH  equal  to  C;  and 
from  the  centre  F,  at  the  dis- 
tance FD,  describe  (3.  Post.) 
the  circle  DKL:  and  from  the  ^ 
centre  G,  at  the  distance  GH, 
describe  (3.  Post.)  another 
circle  HLK;  and  join  KF, 
KG;  the  triangle  KFG  has 
its  sides  equal  to  the  three 
straight  lines  A,  B,  C. 

Because  the  point  F  is  the  centre  of  the  circle  BKL,  FD  is 
equal  (15.  Def.)  to  FK;  but  FD  is  equal  to  the  straight  line  A; 
therefore  FK  is  equal  to  A:  again,  because  G  is  the  centre  of  the 
circle  LKH,  GH  is  equal  (15.  Def.)  to  GK;  but  GH  is  equal  to 
C;  therefore  also  GK  is  equal  to  C;  and  FG  is  equal  to  B;  there- 
fore the  three  straight  lines  KF,  FG,  GH,  are  equal  to  the  three 
A,  B,  C;  and  therefore  the  triangle  KFG  has  its  three  sides  KF, 


Sec  Note. 


BOOK  I.  THE  ELEMENTS  OF  EUCLID.  25 

FG,  GK  equal  to  the  three  given  straight  lines  A,  B,  C.  Which 
was  to  be  done. 

PROP.  XXIII.    PROB. 

At  a  given  point  in  a  given  straight  line,  to  make  a 
rectilineal  angle  equal  to  a  given  rectilineal  angle. 

Let  AB  be  the  given  straight  line,  and  A  the  given  point  in  it, 
and  DCE  the  given  rectilineal  angle;  it  is  required  to  make  an 
angle  at  the  given  point  A  in  C  A 

the  given  straight  line  AB, 
that  shall  be  equal  to  the  given 
rectilineal  angle  DCE. 

Take  in  CD,  CE,  any  points 
D,  E,  and  join  DE,  and  make 
(22.  1.)  the  triangle  AFG,  the 
sides  of  which  shall  be  equal 
to  the  three  straight  lines  D 
CD,  DE,  CE,  so  that  CD  be 
equal  to  AF;  CE  to  AG;  and 
DE  to  FG;  and  because  DC, 

CE  are  equal  to  FA,  AG,  each  to  each,  and  the  base  DE  to  the 
base  FG;  the  angle  DCE  is  equal  (8.  1.)  to  the  angle  FAG. 
Therefore,  at  the  given  point  A  in  the  given  straight  line  AB,  the 
angle  FAG  is  made  equal  to  the  given  rectilineal  angle  DCE. 
Which  was  to  be  done. 


PROP.  XXIV.  THEOR. 

If  two  triangles  have  two  sides  of  the  one  equal  to 
two  sides  of  the  other,  each  to  each,  but  the  angle  con- 
tained by  the  two  sides  of  one  of  them  greater  than 
the  angle  contained  by  the  two  sides  equal  to  them,  of 
the  other;  the  base  of  that  which  has  the  greater  an- 
gle shall  be  greater  than  the  base  of  the  other.* 

Let  ABC,  DEF  be  two  triangles  which  have  the  two  sides  AB, 
AC  equal  to  the  two  DE,  DF,  each  to  each,  viz.  AB  equal  to  DE, 
and  AC  to  DF;  but  the  angle  BAC  greater  than  the  angle  EDF; 
the  base  BC  is  also  greater  than  the  base  EF. 

Of  the  two  sides  DE,  DF,  let  DE  be  the  side  which  is  not 
greater  than  the  other,  and  at  the  point  D,  in  the  straight  line 
DE,  make  (23.  1.)  the  angle  EDG  equal  to  the  angle  BAC;  and 
make  DG  equal  (3.  1.)  to  AC  or  DF,  and  join  EG,  GF. 

Because  AB  is  equal  to  DE,  and  AC  to  DG,  ihc  \\\u  sides  BA, 
AC  are  equal  to  the  two  ED,  DG,  each  to  each,  and  the  angle 

•  See  Note. 
D 


26 


THE  ELEMENTS  OF  EUCLID. 


BOOK  I. 


BAC  is  equal  to  the  angle 
EDG;  therefore  the  base 
BC  is  equal  (4.  1.)  to  the 
base  EG;  and  because 
DG  is  equal  to  DF,  the 
angle  DFG  is  equal  (5.1.) 
to   the   angle   DGF;    but 

the  angle  DGF  is  greater 

than     the     angle     EGF;     '  "     E    "  ^-Z:;       V      ;    Q 

therefore  the  angle  DFG 
is  greater  than  EGF;  and 
much  more  is  the  angle  EFG  greater  than  the  angle  EGF;  and 
because  the  angle  EFG  of  the  triangle  EFG  is  greater  than  its 
angle  EGF,  and  that  the  greater  (19.  1.)  side  is  opposite  to  the 
greater  angle;  the  side  EG  is  therefore  greater  than  the  side  EF; 
but  EG  is  equal  to  BC;  and  therefore  also  BC  is  greater  than 
EF.     Therefore,  if  two  triangles,  Sec.  Q.  E.  D. 

PROP.  XXV.  TilEOR. 

If  two  triangles  have  two  sides  of  the  one  equal  to 
two  sides  of  the  other,  each  to  each,  but  the  base  of 
the  one  greater  than  the  base  of  the  other;  the  angle 
also  contained  by  the  sides  of  that  which  has  the 
greater  base,  shall  be  greater  than  the  angle  contained, 
by  the  sides  equal  to  them,  of  the  other. 

Let  ABC,  DEF  be  two  triangles  which  have  the  two  sides  AB, 
AC  equal  to  the  two  sides  DE,  DF,  each  to  each,  viz.  AB  equal 
to  DE,  and  AC  to  DF;  but  the  base  CB  is  greater  than  the  base 
EF;  the  angle  BAC  is  likewise  greater  than  the  angle  EDF. 

For,  if  it  be  not  greater,  it  must  either  be  equal  to  it,  or  less; 
but  the  angle  BAC  is  not  equal  to  the  angle  EDF,  because  then 
the  base  BC  would  be  equal 
(4.  1.)  to  EF;  but  it  is  not; 
therefore  the  angle  BAC  is  not 
equal  to  the  angle  EDF,  nei- 
ther is  it  less;  because  then 
the  base  BC  would  be  less  (24. 
I.)  than  the  base  EP';  but  it  is 
not;  therefore  the  angle  BAC 
is  not  less  than  t!ie  angle  EDF; 

and  it  was  shown  that  it  is  not  equal  to  it;  therefore  the  angle 
BAC  is  greater  than  the  angle  EDF.  Wherefore,  if  two  trian- 
gles. Sec.  Q.  E.  D. 


PROP.  XXVI.  THEOR. 


If  two  triangles  have  two  angles  of  one  equal  to  two 
angles  of  the  other,  each  to  each;  and  one  side  equal 
to  one  side,  viz.  either  the  sides  adjacent  to  the  equal 


BOOK  I. 


THE  ELEMENTS  OF  EUCLID. 


27 


angles,  or  the  sides  opposite  to  equal  angles  in  each; 
then  shall  the  other  sides  be  equal,  each  to  each ;  and 
also  the  third  angle  of  the  one  to  the  third  angle  of  the 
other. 

Let  ABC,  DEF  be  two  triangles  which  have  the  angles  ABC, 
BCA  equal  to  the  angles  DEF,  EFD,  viz.  ABC  to  DEF,  and  BCA 
to  EFD;  also  one  side  equal  to  one  side;  and  first  let  those  sides 
be    equal    which    are     A  D 

adjacent  to  the  angles     j. 
that  are  equal  in   the  p  In. 
two  triangles,  viz.  BC      i\J\ 
to  EF;  the  other  sides 
shall  be  equal,  each  to 
each,  viz.  AB  to  DE, 
and  AC    to   DF:    and 
the  third   angle  BAC 
to  the  third  angle  EDF.      B  C        E  F 

For,  if  AB  be  not  equal  to  DE,  one  of  them  must  be  the 
greater.  Let  AB  be  the  greater  of  the  two,  and  make  BG  equal 
to  DE  and  join  GC;  therefore,  because  BG  is  equal  to  DE,  and 
BG  to  EF,  the  two  sides  GB,  BC  are  equal  to  the  two  DE,  EF, 
each  to  each;  and  the  angle  GBC  is  equal  to  the  angle  DEF; 
therefore  the  base  GC  is  equal  (4.  1.)  to  the  base  DF,  and  the  tri- 
angle GBC  to  the  triangle  DEF,  and  the  other  angles  to  the  other 
angles,  each  to  each,  to  which  the  equal  sides  are  opposite;  there- 
fore the  angle  GCB  is  equal  to  the  angle  DEE;  but  DFE  is,  by 
the  hypothesis,  equal  to  the  angle  BCA;  wherefore  also  the  angle 
BCG  is  equal  to  the  angle  BCA,  the  less  to  the  greater,  which  is 
impossible;  therefore  AB  is  not  unequal  to  DE,  that  is,  it  is  equal 
to  it,  and  BC  is  equal  to  EF;  therefore  the  two  AB,  BC  ar  •  equul 
to  the  two  DE,  EF,  each  to  each;  and  the  angle  ABC  is  e  ^ual  to 
the  angle  DEF;  the  base  therefore  AC  is  equal  (4.  \.)  to  the  base 
DF,  and  the  third  angle  BAG  to  the  third  angle  EDF. 

Next   let    the     sides 
which    are   oppisite    to    A  D 

equal  angles  in  each  tri- 
angle be  equal  to  one 
another,  viz.  AB  to  DE; 
likewise  in  this  case,  the 
other  sides  shall  be 
equal,  AC  to  DF,  and 
BC  to  EF;  and  also  the 
third  angle  BAC  to  the 
third  EDF.  B  H     C       E  F 

For,  if  BC  be  not  equal  to  EF,  let  BC  be  the  greater  of  them, 
and  make  BH  equal  to  EF,  and  join  AH;  and  because  BH  is  equal 
to  EF,  and  AB  to  DE,  the  two  AB,  BH  are  equal  to  the  two  DE, 
EF  each  to  each;  and  they  contain  equal  angles;  therefore  the 
base  AH  is  equal  to  the  base  DF,  and  the  triangle  ABH  to  the 
triangle  DEF,  and  the  other  angles  shall  be  equal,  each  to  each, 
.  to  which  the  equal  sides  are  opposite;  therefore  the  angle  BHA 


28 


THE  ELEMENTS  OF  EUCLID. 


BOOK  I. 


is  equal  to  the  angle  EFD;  but  EFD  is  equal  to  the  angle  BCA5 
therefore  also  the  angle  BHA  is  equal  to  the  angle  BCA,  that  is, 
the  exterior  angle  BHA  of  the  triangle  AHC  is  equal  to  its  in- 
terior and  opposite  angle  BCA;  which  is  impossible;  (16.  1.) 
wherefore  BC  is  not  unequal  to  EF,  that  is,  it  is  equal  to  it;  and 
AB  is  equal  to  DE;  therefore  the  two  AB,  BC  are  equal  to  the 
two  DEj  EF,  each  to  each;  and  they  contain  equal  angles;  where- 
fore the  base  AC  is  equal  to  the  base  DF,  and  the  third  angle 
BAC  to  the  third  angle  EDF.  Therefore,  if  two  triangles,  &c. 
Q.  E.  D. 

PROP.  XXVII.  THEOR. 

If  a  straight  line  falling  upon  two  other  straight 
lines  makes  the  alternate  angles  equal  to  one  another, 
these  two  straight  lines  shall  be  parallel. 

Let  the  straight  line  EF,  which  falls  upon  the  two  straight 
lines  AB,  CD  make  the  alternate  angles  AEF,  EFD  equal  to  one 
another;  AB  is  parallel  to  CD. 

For,  if  it  be  not  parallel,  AB  and  CD  being  produced  shall  meet 
either  towards  B,  D,  or  towards  A,  C;  let  them  be  produced  and 
meet  towards  B,  D,  in  the  point  G;  therefore  GEF  is  a  triangle, 
and  its  exterior  angle  AEF  is  greater  (16.  1.)  than  the  interior 
and  opposite  angle  EFG;  but 
it  is  also  equal  to  it,  which  is 
impossible;  therefore  AB  and  * 
CD  being  produced  do  not 
meet  towards  B,  D.  In  like 
manner  it  may  be  demonstrat-  C 
ed  that  they  do  not  meet  to- 
wards A,  C;  but  those  straight 
lines  which  meet  neither  way, 

though  produced  ever  so  far,  are  parallel  (35.  def.)  to  one  another. 
AB  therefore  is  parallel  to  CD.  Wherefore,  if  a  straight  line, 
&c.  Q.  E.  D. 

PROP.  XXVIII.  THEOR. 

If  a  straight  line  falling  upon  two  other  straight 
lines  makes  the  exterior  angle  equal  to  the  interior  and 
opposite  upon  the  same  side  of  the  line;  or  makes  the 
interior  angles  upon  the  same  side  together  equal  to 
two  right  angles ;  the  two  straight  lines  shall  be  paral- 
lel to  one  another. 

Let  the  straight  line  EF,  which  falls  E 
upon  the  two  straight  lines  AB,  CD,  ^. 
make  the  exterior  angle  EGB  equal  to  \  q 

the  interior  and  opposite  angle  GHD  A -^^ '  B 

upon  the  same  side;  or  make  the  inte- 
rior angles  on  the  same  side   BGH, 

GHD  together  equal  to  two  right  an-  C _„ \ U 

gles;  AB  is  parallel  to  CD.  H  \ 

Because  the  angle  EGB  is  equal  to  ^ 

F 


BOOK  I.  THE    ELEMENTS    OF  EUCLID.  29 

the  angle  GHD,  and  the  angle  EGB,  equal  (15.  1.)  to  the  angle 
AGH,  the  angle  AGH  is  equal  to  the  angle  GHD;  and  they  are 
the  alternate  angles;  therefore  AB  is  parallel  (27.  1.)  to  CD. 
Again,  because  the  angles  BGH,  GHD  are  equal  (by  hyp.)  to  two 
right  angles;  and  that  AGH,  BGH  are  also  equal  (l3.  1.)  to  two 
right  angles;  the  angles  AGH,  BGH  are  equal  to  the  angles 
BGH,  GHD:  take  away  the  common  angle  BGH;  therefore  the 
remaining  angle  AGH  is  equal  to  the  remaining  angle  GHD; 
and  they  are  alternate  angles:  therefore  AB  is  parallel  to  CD. 
Wherefore,  if  a  straight  line,  8cc.  Q.  E.  D. 

PROP.  XXIX.  THEOR. 

If  a  straight  line  fall  upon  two  parallel  straight  lines, 
it  makes  the  alternate  angles  equal  to  one  another;  and 
the  exterior  angle  equal  to  the  interior  and  opposite 
upon  the  same  side;  and  hkewise  the  two  interior  an- 
gles upon  the  same  side  together  equal  to  two  right 
angles.^ 

Let  the  straight  line  EF  fall  upon  the  parallel  straight  lines 
AB,  CD;  the  alternate  angles,  AGH,  GHD  are  equal  to  one  ano- 
ther; and  the  exterior  angle  EGB  is  equal  to  the  interior  and  op- 
posite upon  the  same  side  GHD,  E 
and  the  two  interior  angles  BGH, 
GHD   upon  the  same  side  are  to- 
gether equal  to  two  right  angles.       » 

For  if  AGH  be  not  equal  to 
GHD,  one  of  them  must  be  great- 
er than  the  other;  let  AGH  be  the  p  \G  P^ 
greater;  and  because  the  angle 
AGH  is  greater  than  the  angle 
GHD,  add  to  each  of  them  the  an- 
gle BGH,  therefore  the  angles  AGH,  F 
BGH  are  greater  than  the  angles  BGH,  GHD;  but  the  angles 
AGH,  BGH  are  eqaul  (13.  1.)  to  two  right  angles;  therefore 
the  angles  BGH,  GHD  are  less  than  two  right  angles;  but  those 
straight  lines  which,  with  another  straight  line  falling  upon  them, 
make  the  interior  angles  on  the  same  side  less  than  two  right  an- 
gles, do  meet  (12.  ax.)*  together  if  continually  produced;  there- 
fore the  straight  lines  AB,  CD,  if  produced  far  enough,  shall  meet; 
but  they  never  meet,  since  they  are  parallel  by  the  hypothesis; 
therefore  the  angle  AGH  is  not  unequal  to  the  angle  GHD,  that 
is,  it  is  equal  to  it;  but  the  angle  AGH  is  equal  (l5.  1.)  to  the  an- 
gle EGB;  therefore  likewise  EGB  is  equal  to  GHD;  add  to  each 
of  these  the  angle  BGH;  therefore  the  angles  EGB,  BGH  are 
equal  to  the  angles  BGH,  GHD;  but  EGB,  BGH  are  equal  (13.  1.) 
to  two  right  angles;  therefore  also  BGH,  GHD  are  equal  to  two 
right  angles.     Wherefore,  if  a  straight  line.  Sec.  Q.  E.  D. 

"  See  the  Notes  to  this  Proposition. 


B 


30  THE  ELEMENTS  OF  EUCLID.  BOOK  I. 


PROP.  XXX.  THEOR. 

Straight  lines  which  are  parallel  to  the  same  straight 
line  are  parallel  to  one  another. 

Let  AB,  CDj  be  each  of  them  parallel  to  EF,  AB  is  also  paral- 
lel to  CD. 

Let  the  straight  line  GHK  cut  AB,  EF,  CD;  and  because  GHK 
cuts  the  parallel  straight  lines  AB,  EF, 
the  angle  AGH  is  equal  (29.  L)  to  the 
angle  GHF.    Again,  because  the  straight  » 
line  GK  cuts  the  parallel  straight  lines 
EF,  CD,  the  angle  GHF  is  equal  to  (29. 
L)  the  angle  GKD;  and  it  was  shown  -p 
that  the  angle  AGK  is  equal  to  the  an- 
gle GHF;  therefore  also  AGK  is  equal  p 
to  GKD;  and  they  are  alternate  angles; 
therefore  AB  is  parallel  (27.  1.)  to  CD. 
Wherefore  straight  lines,  Sec.  Q.  E.  D. 


PROP.  XXXL   PROB. 

To  draw  a  straight  line  through  a  given  point  pa- 
rallel to  a  given  straight  line. 

Let  A  be  the  given  point,  and  BC  the  given  straight  line;  it  is 

required  to  draw  a  straight  line  through  E A F 

the  point  A,  parallel  to  the  straight  line  ~7 

BC.  /        f 

In  BC  take  any  point  D,  and  join  X 

AD;  and  at  the  point  A,  in  the  straight  ^ ^^ p 

line  AD,  make  (23.  L)  the  angle  DAE  ^  ^  ^ 

equal  to  the  angle  ADC;  and  produce  the  straight  line  EA  to  F. 

Because  the  straight  line  AD,  which  meets  the  two  straight 
lines  BC,  EF,  makes  the  alternate  angles  EAD,  ADC  equal  to  one 
another,  EF  is  parallel  (27.  1.)  to  BC.  Therefore  the  straight 
line  EAF  is  drawn  through  the  given  point  A  parallel  to  the 
given  straight  line  BC.     Which  was  to  be  done. 


PROP.  XXXIL  TFIEOR. 

If  a  side  of  any  triangle  be  produced,  the  exterior 
angle  is  equal  to  the  two  interior  and  opposite  angles ; 
and  the  three  interior  angles  of  every  triangle  are  equal 
to  two  right  angles. 

Let  ABC  be  a  triangle,  and  let  one  of  its  sides  BC  be  produced 
to  D;  the  exterior  angle  ACD  is  equal  to  the  two  interior  and  op- 
posite angles  CAB,  ABC;  and  the  three  interior  angles  of  the 


BOOK  I. 


THE  ELEMENTS  OF  EUCLID. 


31 


D 


triangle,  viz.  ABC,  BCA,  CAB,  are  together  equal  to  two  right 
angles. 

Through  the  point  C  draw  A 

CE  parallel  (31.    1.)  to  the  y\  ^  F, 

straight  line  AB^  and  be- 
cause AB  is  parallel  to  CE 
and  AC  meets  them,  the  al- 
ternate angles  BAC,  ACE 
are   equal  (29.    1.)     Again; B  C  D 

because  A  is  parallel  to  CE,  and  BD  falls  upon  them,  the  exte- 
rior angle  ECD  is  equal  to  the  interior  and  opposite  angle  ABC; 
but  the  angle  ACE  was  shown  to  be  equal  to  the  angle  BAC; 
therefore  the  whole  exterior  angle  ACD  is  equal  to  the  two  inte- 
rior and  opposite  angles  CAB,  ABC;  to  these  equals  add  the  an- 
gle ACB,  and  the  angles  ACD,  ACB  are  equal  to  the  three  an- 
gles CBA,  BAC,  ACB;  but  the  angles  ACD,  ACB  are  equal  (13. 
1.)  to  two  right  angles:  therefore  also  the  angles  CBA,  BAC, 
ACB  are  equal  to  two  right  angles.  Wherefore,  if  a  side  of  a 
triangle.  Sec.  Q.  E.  D. 

Cor.  1.  All  the  interior  angles  of  any- 
rectilineal  figure,  together  with  four  right 
angles,  are  equal  to  twice  as  many  right  E 
angles  as  the  figure  has  sides. 

For  any  rectilineal  figure  ABCDE  can 
be  divided  into  as  many  triangles  as  the 
figure  has  sides,  by  drawing  straight  lines 
from  a  point  F  within  the  figure,  to  each 
of  its  angles.  And,  by  the  preceding  pro- 
position, all  the  angles  of  these  triangles 
are  jqual  to  twice  as  many  right  angles  as  there  are  triangles, 
thatis,  as  there  are  sides  of  the  figure;  and  the  same  angles  are 
equal  to  the  angles  of  the  figure,  together  with  the  angles  at  the 
point  F,  which  is  the  common  vertex  of  the  triangles:  that  is, 
(2  Cor.  15.  1.)  together  with  four  right  angles.  Therefore  all  the 
angles  of  the  figure,  together  with  four  right  angles,  are  equal  to 
twice  as  many  right  angles  as  the  figure  has  sides. 

Cor.  2.  All  the  exterior  angles  of  any  rectilineal  figure,  are  to- 
gether equal  to  four  right  angles. 

Because  every  interior  angle  ABC, 
with  its  adjacent  exterior  ABD,  is 
equal  (13.  1.)  to  two  right  angles: 
therefore  all  the  interior,  together 
with  all  the  exterior  angles  of  the 
figure,  are  equal  to  twice  as  many 
right  angles  as  there  are  sides  of  the 
figure;  that  is,  by  the  foregoing  co-D, 
rollary,  they  are  equal  to  all  the  in- 
terior angles  of  the  figure,  together 
with  four  right  angles;  therefore  all  the  exterior  angles  are  equal 
to  four  right  angles. 


32  THE  ELEMENTS  OF  EUCLID.  BOOK  I. 

PROP.  XXXm.  THEOR. 

The  straight  lines  which  join  the  extremities  of  two 
equal  and  parallel  straight  lines,  towards  the  same 
parts,  are  also  themselves  equal  and  parallel. 

Let   AB,    CD   be   equal    and    parallel  A  B 

straight   lines,    and    joined    towards    the 
same  parts  by  the  straight  lines  AC, 
AC,  BD  are  also  equal  and  parallel. 

Join  BC;  and  because  AB  is  parallel  to 

CD;   and  BC  meets  them,  the   alternate  

angles  ABC,  BCD  are  equal  (29.  1.);  and      ^  D 

because  AB  is  equal  to  CD,  and  BC  common  to  the  two  triangles 
ABC,  DCB,  the  two  sides  AB,  BC  are  equal  to  the  two  DC,  CB; 
and  the  angle  ABC  is  equal  to  the  angle  BCD;  therefore  the  base 
AC  is  equal  (4.  1.)  to  the  base  BD,  and  the  triangle  ABC  to  the 
triangle  BCD,  and  the  other  angles  to  the  other  angles,  (4.  1.) 
each  to  each,  to  which  the  equal  sides  are  opposite:  therefore  the 
angle  ACB  is  equal  to  the  angle  CBD;  and  because  the  straight 
line  BC  meets  the  two  straight  lines  AC,  BD,  and  makes  the  al- 
ternate angles  ACB,  CBD  equal  to  one  another,  AC  is  parallel 
(27.  1.)  to  BD;  and  it  was  shown  to  be  equal  to  it.  Therefore 
straight  lines,  &c.  Q.  E.  D. 

PROP.  XXXIV.  THEOR. 

The  opposite  sides  and  angles  of  parallelograms  are 
equal  to  one  another,  and  the  diameter  bisects  them, 
that  is,  divides  them  into  two  equal  parts.  # 

N.  B.  ./?  jmrailelogram  is  a  four  sided  figure^  of  which  the  oppo- 
site sides  are  parallel;  and  the  diameter  is  the  straight  line  joining 
two  of  its  opposite  angles. 

Let  ACDB  be  a  parallelogram,  of  which  BC  is  a  diameter;  the 
opposite  sides  and  angles  of  the  figure  are  equal  to  one  another; 
and  the  diameter  BC  bisects  it. 

Because  AB  is  parallel  to  CD,  and 
BC  meets  them,  the  alternate  angles  r 
ABC,  BCD  are  equal  (29.  1.)  to  one  ano- 
ther; and  because  AC  is  parallel  to  BD, 
and  BC  meets  them,  the  alternate  angles 
ACB,  CBD  are  equal  (29.  1.)  to  one  ano- 
ther; wherefore  the  two  triangles  ABC,  C  D' 
CBD  have  two  angles  ABC,  BCA  in  one,  equal  to  two  angles 
BCD,  CBD  in  the  other,  each  to  each,  and  one  side  BC  common 
to  the  two  triangles,  which  is  adjacent  to  their  equal  angles; 
therefore  their  other  sides  shall  be  equal,  each  to  each,  and  the 
third  angle  of  the  one  to  the  third  angle  of  the  other,  (26.  1.) 
viz.  the  side  AB  to  the  side  CD,  and  AC  to  BD,  and  the  angle 
BAC  equal  to  the  angle  BDC;  and  because  the  angle  ABC  is 


BOOK  I.  THE  ELEMENTS  OF  EUCLID,  33 

equal  to  the  angle  BCD,  and  the  angle  CBD  to  the  angle  ACB, 
the  whole  angle  ABD  is  equal  to  the  whole  angle  ACD:  and  the 
angle  BAC  has  been  shown  to  be  equal  to  the  angle  BDC;  there- 
fore the  opposite  sides  and  angles  of  parallelograms  are  equal  to 
one  another;  also,  their  diameter  bisects  them;  for  AB  being 
equal  to  CD,  and  BC  common,  the  two  AB,  BC  are  equal  to  the 
two  DC,  CB,  each  to  each;  and  the  angle  ABC  is  equal  to  the 
angle  BCD;  therefore  the  triangle  ABC  is  equal  (4.  1.)  to  the 
triangle  BCD,  and  the  diameter  BC  divides  the  parallelogram 
ACDB  into  two  equal  parts.  Q.  E.  D. 

PROP.  XXXV.  THEOR. 

Parallelograms  upon  tlie  same  base,  and  between 
the  same  parallels,  are  equal  to  one  another*^ 

Let  the  parallelograms  ABCD,  EBCF  be  upon  the  same  base 
BC,  and  betw^een  the  same  parallels  AF,  BC;  the  parallelogram 
ABCD  shall  be  equal  to  the  parallelogram  EBCF.f 

If  the  sides  AD,  DF  of  the  parallelo-  AD  F 

grams  ABCD,  DBCF  opposite  to  the  base 
BC  be  terminated  in  the  same  point  D;  it 
is  plain  that  each  of  the  parallelograms  is 
double  (34.  1.)  of  the  triangle  BDC;  and 
they  are  therefore  equal  to  one  another. 

But,  if  the  sides  AD,  EF,  opposite  tojj  q 

the  base  BC  of  the  parallelograms  ABCD, 

EBCF,  be  not  terminated  in  the  same  point;  then,  because  ABCD 
is  a  parallelogram,  AD  is  equal  (34.  1.)  to  BC;  for  the  same  rea- 
son EF  is  equal  to  BC;  wherefore  AD  is  equal  (l.  Ax.)  to  EF: 
and  DE  is  common;  therel^ore  the  whole,  or  the  remainder  AE, 
is  equal  (2.  or  3.  Ax.)  to  the  whole,-  or  the  remainder  DF;  AB 
also  is  equal  to  DC;  and  the  two  EA,  AB  are  therefore  equal  to 

A  D     E  F       A       E         D        F 


/ 

/ 

/ 

/ 

/ 

/ 

1/ 

/ 

/ 

\  I    \ 

\  i      \ 


\ 


\ 


t 


B  C  B  C 

the  two  FD,  DC,  each  to  each;  and  the  exterior  angle  FDC  is 
equal  (19.  1.)  to  the  interior  EAB;  therefore  the  base  EB  is  equal 
to  the  base  FC,  and  the  triangle  EAB  equal  (4.  1.)  to  the  triangle 
FDC;  take  the  triangle  FDC  from  the  trapezium,  ABCF,  and 
from  the  same  trapezium  take  the  triangle  EAB;  the  remainders 
therefore  are  equal,  (3.  Ax.)  that  is,  the  parallelogram  ABCD  is 
equal  to  the  parallelogram  EBCF.  Therefore,  parallelograms 
upon  the  same  base,  &c.  Q,  E.  D. 

*'  See  Note.  t  See  the  2d  nnd  3d  figures. 

E 


34 


THE  ELEMENTS  OF  EUCLID. 


BOOK  I. 


PROP.  XXXVI.  THEOR. 

Parallelograms  upon  equal  bases,  and  between  the 
same  parallels,  are  equal  to  one  another. 

Let  ABCD,  EFGH  be  pa-  A  D  E  H 

rallelog'rams  upon  equal  ba- 
ses BC,  FG,  and  between  the 
same  parallels  AH,  BG;  the 
parallelogram  ABCD  is  equal 
to  EFGH. 

Join  BE,  CH;  and  because 

BC  is  equal  to  FG,  and  FG      B  C      F  G 

to  (34.  1.)  EH,  BC  is  equal  to  EH;  and  they  are  parallels,  and 
joined  towards  the  same  parts  by  the  straight  lines  BE,  CH;  but 
straight  lines  which  join  equal  and  parallel  straight  lines  towards 
the  same  parts,  are  themselves  equal  and  parallel;  (33.  1)  there- 
fore EB,  CH  are  both  equal  and  parallel,  and  EBCH  is  a  parallel- 
ogram; and  it  is  equal  (35.  1.)  to  ABCD,  because  it  is  upon  the 
same  base  BC,  and  between  the  same  parallels  BC,  AD:  for  the 
like  reason,  the  parallelogram  EFGH  is  equal  to  the  same  EBCH; 
therefore  also  the  parallelogram  ABCD  is  equal  to  EFGH. 
Wherefore  parallelograms,  &c.  Q.  E.  D. 

PROP.  XXXVH.  THEOR. 

Triangles  upon  the  same  base,  and  between  the 
same  parallels,  are  equal  to  one  another. 

Let  the  triangles  ABC,  DBC  be  upon  the  same  base  BC,  and 
between  the  same  parallels  AD,  E  AD  F 

BC:  the  triangle  ABC  is  equal 
to  the  triangle  DBC. 

Produce  AD  both  ways  to  the 
points  E,  F,  and  through  B  draw 
(31.  1.)  BE  parallel  to  CA;  and 
through    C    draw    CF    parallel  ^^. 

to    BD:    therefore   each  of  the  B  C 

figures  EBCA,  DBCF  is  a  parallelogram;  and  EBCA  is  equal 
(35.  1.)  to  DBCF,  because  they  are  upon  the  same  base  BC,  and 
between  the  same  parallels  BC,  EF;  and  the  triangle  ABC  is  the 
half  of  the  parallelogram  EBCA,  because  the  diameter  AB  bisects 
(34.  1.)  it;  and  the  triangle  DBC  is  the  half  of  the  parallelogram 
DBCF,  because  the  diameter  DC  bisects  it:  but  the  halves  of 
equal  things  are  equal:  (7.  Ax.)  therefore  the  triangle  ABC  is 
equal  to  the  triangle  DBC.     Wherefore  triangles,  &c.  Q.  E.  D. 

PROP.  XXXVHL  THEOR. 

Triangles  upon  equal  bases,  and  between  the  same 
parallels,  are  equal  to  one  another. 

Let  the  triangles  ABC,  DEF  be  upon  equal  bases  BC,  EF,  and 


BOOK  I. 


THE  ELEMENTS  OF  EUCLID. 


35 


between  the  same  parallels  BF,  AD:  the  triangle  ABC  is  equal  to 
the  triangle  DEF. 

Produce  AD  both  ways  to  the  points  G,  H,  and  through  B  draw 
BG  parallel  (31.  1.)  to  CA,  and  through  F  draw  FH  parallel  to 
ED:  then  each  of  the 
figures  GBCA,  DEFH  is 
a  parallelogram;  and  they 
are  equal  (36.  1.)  to  one 
another,  because  they  are 
upon  equal  bases  BC,  EF, 
and  between  the  same  pa- 
rallels BF,  GH;  and  the 
triangle  ABC  is  the  half 
(34.  1.)  of  the  parallelogram  GBCA,  because  the  diameter  AB 
bisects  it;  and  the  triangle  DEF  is  the  half  (34.  1.)  of  the  parallel- 
ogram DEFH,  because  the  diameter  DF  bisects  it:  but  the  halves 
of  equal  things  are  equal;  (7.  Ax.)  therefore  the  triangle  ABC  is 
equal  to  the  tiiangle  DEF.     Wherefore  triangles,  Sec.  Q.  E.  D. 

PROP.    XXXIX.  THEOR. 

Equal  triangles  upon  the  same  base,  and  upon  the 
same  side  of  it,  are  between  the  same  parallels. 

Let  the  equal  triangles  ABC,  DBC  be  upon  the  same  base  BC, 
and  upon  the  same  side  of  it;  they  are  between  the  same  parallels. 

Join  AD;  AD  is  parallel  to  BC:  for,  if  it  is  not,  through  the 
point  A  draw  (31.  1.)  AE  parallel  to  BC,  and  join  EC;  the  trian- 
gle ABC  is  equal  (37.  1.)  to  the  triangle  EBC,  because  it  is  upon 
the  same  base  BC,  and  between  the  same     A  D 

parallels  BC,  AE:  but  the  triangle  ABC  is 
equal  to  the  triangle  BDC;  therefore  also 
the  triangle  BDC  is  equal  to  the  triangle 
EBC,  the  greater  to  the  less,  which  is  im- 
possible; therefore  AE  is  not  parallel  to 
BC.  In  the  same  manner,  it  can  be  de- 
monstrated that  no  other  line  but  AD  is  B  C 
parallel  to  BC:  AD  is  therefore  parallel  to  it.  Wherefore  equal 
triangles  upon,  Sec.  Q.  E.  D. 

PROP.  XL.  THEOR. 

Equal  triangles  upon  equal  bases,  in  the  same 
straight  line,  and  towards  the  same  parts,  are  between 
the  same  parallels. 

Let  the  equal  triangles  ABC,  DEF  be  upon  equal  bases  BC, 
EF,  in  the  same  straight  line  '^  ^ 

BF,  and  towards  the  same  parts; 
they  are  between  the  same  pa- 
rallels. 

Join  AD;  AD  is  parallel  to 
BC;  for. if  it  is  not,  through  A 
draw  (31.  1.)  AG  parallel  to 
BF,  and  join  GF:  the  triangle  B 


D 


lllE    ELEMENTS   OF   EUCLID. 


BOOK  I. 


D 


ABC  is  equal  (38.  I.)  to  the  triangle  GEF,  because  they  are 
upon  equal  bases  BC,  EF,  and  between  the  same  parallels  BE, 
AG :  but  the  trians^le  ABC  is  equal  to  the  triangle  DEF5  therefore 
also  the  triangle  DEE  is  equal  to  the  triangle  GEE,  the  greater 
to  the  less,  which  is  impossible;  therefore  AG  is  not  parallel  to 
BE:  and  in  the  same  manner  it  can  be  demonstrated  that  there 
is  no  other  parallel  to  it  but  AD;  AD  is  therefore  parallel  to 
BE.     Wherefore,  equal  triangles,  S<:c.  Q.  E.  D. 

PROP.  XLI.  THEOR. 

If  a  parallelogram  and  triangle  be  upon  the  same 
base,  and  between  the  same  parallels;  the  parallelogram 
shall  be  double  of  the  triangle. 

Let  the  parallelogram  ABCD  and  the  triangle  EBC  be  upon 
the  same  base  BC,  and  between  the  same  parallels  BC,  AE;  the 
parallelogram  ABCD  is  double  of  the  triangle  EBC. 

Join  AC;  then  the  triangle  ABC  is  A 
equal  (S7.  1.)  to  the  triangle  EBC,  be- 
cause they  are  upon  the  same  base  BC, 
and  between  the  same  parallels  BC,  AE. 
But  the  parallelogram  ABCD  is  double 
(34.  1.)  of  the  triangle  ABC,  because  the 
diameter  AC  divides  it  into  two  equal 
parts;  wherefore  ABCD  is  also  double 
of  the  triangle  EBC.  Therefore  if  a  pa- 
rallelogram, Sec.  Q.  E.   D. 

PROP.  XLII.  PROB. 

To  describe  a  parallelogram  that  shall  be  equal  to  a 
given  triangle,  and  have  one  of  its  angles  equal  to  a 
given  rectilineal  angle. 

Let  ABC  be  the  given  triangle,  and  D  the  given  rectilineal  an- 
gle. It  is  required  to  describe  a  parallelogram  that  shall  be  equal 
to  the  giv6n  triangle  ABC,  and  have  one  of  its  angles  equal  to  D. 

Bisect  (10.  1.)  BC  in  E,  join  AE,  and  at  the  point  E  in  the 
straight  line  EC  make  (23.  1.)  the  angle  CEF  equal  to  D;  and 
through  A  draw  (31.  1.)  AG  parallel  to  EC,  and  through  C  draw 


A     E 


G 


CG  (31.1.)  parallel  to  EF:  therefore 
EECG  is  a  parallelogram:  and  be- 
cause BE  is  equal  to  EC,  the  trian- 
gle ABE  is  likewise  equal  (38.  1.)  to 
the  triangle  AEC,  since  they  are  upon 
equal  bases  BE,  EC,  and  between 
Uie  same  parallels  BC,  AG;  there- 
fore the  triangle  ABC  is  double  of 
the  triangle  AEC;  and  the  parallel- 
ogram EECG  is  likewise  double  g 
(41.  1.)  of  the  triangle  AEC,  be- 
cause it  is  upon  the  same  base,  and  between  the  same  parallels: 


BOOK  I. 


THE  ELKMENTS  OF  EUCLID. 


therefoi'e  the  parallelogram  FECG  is  equal  to  the  triangle  ABC, 
and  it  has  one  of  its  angles  CEF  equal  to  the  given  angle  D. 
Wherefore  there  has  been  described  a  parallelogram  FECG  equal 
to  a  given  triangle  ABC,  having  one  of  its  angles  CEF  equal  to 
the  given  angle  D.     Which  was  to  be  done. 

PROP.  XLIII.  THEOR. 

The  compliments  of  the  parallelograms  which  are 
about  the  diameter  of  any  parallelogram,  are  equal  to 
one  another. 

Let  ABCD  be  a  parallelogram,  of  which  the  diameter  is  AC, 
and  EH,  FG  the  parallelograms  about       AH  D 

AC,  that  is^  through  which  AC  passes,        \\         j~j' 
and  BK,  KD  the  other  parallelograms       /    ^x  /  ^ 
which    make    up    the    whole    figure  EJ  — 
ABCD,   which    arc    therefore    called 
the    complements:    the    complement 
BK  is  equal  to  the  complement  KD. 

Because  ABCD  is  a  parallelogram, 
and  AC  its  diameter,  the  triangle 
ABC  is  equal  (34.  1.)  to  the  triangle  ■" 
ADC;  and  because  EKHA  is  a  parallelogram,  the  diameter  of 
which  is  AK,  the  triangle  AEK  is  equal  to  the  triangle  AHK:  by 
the  same  reason,  the  triangle  KGC  is  equal  to  the  triangle  KFC: 
then,  because  the  triangle  AEK  is  equal  to  the  triangle  AHK,  and 
the  triangle  KGC  to  KFC;  the  triangle  AEK,  together  wdth  the 
triangle  KGC,  is  equal  to  the  triangle  AHK  together  with  the 
triangle  KFC:  but  the  whole  triangle  ABC  is  ecjual  to  the  whole 
ADC;  therefore  the  remaining  complement  BK  is  equal  to  the 
remaining  complement  KD.  Wherefore  the  complements,  &c. 
Q.  E.  D. 

PROP.  XLIV.    PROB. 

To  a  given  straight  line  to  apply  a  parallelogram, 
which  shall  be  equal  to  a  given  triangle,  and  have  one 
of  its  angles  equal  to  a  given  rectilineal  angle. 

Let  AB  be  the  given  straight  line,  and  C  the  given  triangle, 
and  D  the  given  rectilineal  angle.  It  is  required  to  apply  to  the 
straight  line  AB  a  parallelogram  equal  to  the  triangle  C,  and 
having  an  angle  equal  to  D. 

Make  (42.  1.) 
the  parallelogram 
BEFG  equal  to 
the  triangle  C, and 
having  the  angle 
EBG  equal  to  the 
angle  D,  so  that 
BE  be  in  the  same 
straight  line  with 
AB,  and  produce 
FG    to    H;     and 


38  THE  ELEMENTS  OF  EUCLID.  BOOK  I. 

through  A  draw  (31.  1.)  AH  parallel  to  BG  or  EF,  and  join  HB. 
Then,  because  the  straight  line  HF  falls  upon  the  parallels  AH, 
EF,  the  angles  x\HF,  HFE  are  together  equal  (29.  1.)  to  two 
right  angles:  wherefore  the  angles  BHF,  HFE  are  less  than  two 
right  angles:  but  straight  lines  which  with  another  straight  line 
make  the  interior  angles  upon  the  same  side  less  than  two  right 
angles,  do  meet  (12.  Ax.)  if  produced  far  enough:  therefore  HB, 
FE  dhall  meet,  if  produced;  let  them  meet  in  K,,  and  through  K 
draw  KL  parallel  to  EA  or  FH,  and  produce  HA,  GB  to  the 
points  LM:  then  HLKF  is  a  parallelogram,  of  which  the  diame- 
ter is  HK,  and  AG,  ME  are  the  parallelograms  about  HK;  and 
LB,  BF  are  the  complements;  therefore  LB  is  equal  (43.  1.)  to 
BF;  but  BF  is  equal  to  the  triangle  C:  wherefore  LB  is  equal  to 
the  triangle  C:  and  because  the  angle  GBE  is  equal  (15.  l.)to 
the  angle  ABM,  and  likewise  to  the  angle  D;  the  angle  ABM  is 
equal  to  the  angle  D:  therefore  the  parallelogram  LB  is  applied 
to  the  straight  line  AB,  is  equal  to  the  triangle  C,  and  has  the 
angle  ABM  equal  to  the  angle  D.     Which  was  to  be  done. 

PROP.  XLV.    PROB. 

To  describe  a  parallelogram  equal  to  a  given  recti- 
lineal figure,  and  having  an  angle  equal  to  a  given  rec- 
tilineal angle."^ 

Let  ABCD  be  the  given  rectilineal  figure,  and  E  the  given  rec- 
tilineal angle.  It  is  required  to  describe  a  parallelogram  equal 
to  ABCD,  and  having  an  angle  equal  to  E. 

Join  DB,  and  describe  (42.  1.)  the  parallelogram  FH  equal  to 
the  triangle  ADB,  and  having  the  angle  HKF  equal  to  the  angle 
E;  and  to  the  straight  line  GH  apply  (44.  1.)  the  parallelogram 
GM  equal  to  the  triangle  DBC,  having  the  angle  GHM  equal  to 
the  angle  E;  and  because  the  angle  E  is  equal  to  each  of  the  an- 
gles FKH,  GHM,  the  angle  FKH  is  equal  to  GHM:  add  to  each 
of  these  the  angle  KHG;  therefore  the  angles  FKH,  KHG,  are 
equal   to    the    angles  A  D  F         G         L 

KHG,  GHM;  but 
FKH,  KHG  are  equal 
(29.  1.)  to  two  right 
angles:  therefore  also 
KHG,  GHM  are  equal 
to  two  right  angles; 
and  because  at  the 
point  H  in  the  straight 
line     GH,     the     two  ^  C       K         H        M 

straight  lines  KH,  HM,  upon  the  opposite  sides  of  it,  make  the 
adjacent  angles  equal  to  two  right  angles,  KH  is  in  the  same 
straight  line  (14.  1.)  with  HM,  and  because  the  straight  line  HG 
meets  the  parallels  KM,  FG;  the  alternate  angles  MHG,  HGF 
are  equal:  (29.  1.)  add  to  each  of  these  the  angle  HGL;  therefore 

*  See  Note. 


BOOK  I.  THE  ELEMENTS  OF  EUCLID.  *  39 

the  angles  MHG,  HGL  are  equal  to  the  angles  HGF,  HGL;  but 
the  angles  HGM,  HGL  are  equal  (29.  1.)  to  two  right  angles; 
wherefore  also  the  angles  HGF,  HGL  are  equal  to  two  right  an- 
gles, and  FG  is  therefore  in  the  same  straight  line  with  GL:  and 
because  KF  is  parallel  to  HG,  and  HG  to  ML;  KF  is  parallel 
(30.  1.)  to  ML;  and  KM,  FL  are  parallels;  wherefore  KFLM  is 
a  parallelogram;  and  because  the  triangle  ABD  is  equal  to  the 
parallelogram  HF,  and  the  triangle  DBC  to  the  parallelogram 
GM;  the  whole  rectilineal  figure  ABCD  is  equal  to  the  whole 
parallelogram  KFLM;  therefore  the  parallelogram  KFLM  has 
been  described  equal  to  the  given  rectilineal  figure  ABCD,  having 
the  angle  FKM  equal  to  the  given  angle  E.  Which  was  to  be 
done. 

Cor.  From  this  it  is  manifest  how  to  a  given  straight  line  to 
apply  a  parallelogram,  which  shall  have  an  angle  equal  to  a  given 
rectilineal  angle,  and  shall  be  equal  to  a  given  rectilineal  figure, 
viz.  by  applying  (44.  1.)  to  the  given  straight  line  a  parallelogram 
equal  to  the  first  triangle  ABD,  and  having  an  angle  equal  to  the 
given  angle. 

PROP.  XLVL   PROB. 
To  describe  a  square  upon  a  given  straight  line. 

Let  AB  be  the  given  straight  line;  it  is  required  to  describe  a 
square  upon  AB. 

From  the  point  A  draw(ll.  1.)  AC  at  right  angles  to  AB;  and 
make  (3.  1.)  AD  equal  to  AB,  and  through  the  point  D  draw  DE 
parallel  (31.  1.)  to  AB,  and  through  B  draw  BE  parallel  to  AD; 
therefore  ADEB  is  a  parallelogram:  whence  AB  is  equal  (34.  1.) 
to  DE,  and  AD  to  BE;  but  BA  is  equal  to  C 
AD;  therefore,  the  four  straight  lines  BA, 
A'D,  DE,  EB  are  equal  to  one  another,  and 
the  parallelogram  ADEB  is  equilateral, 
likewise  all  its  angles  are  right  angles: 
because  the  straight  lines  AD  meeting  the 
parallels  AB,  DE,  the  angles  BAD,  ADE  D 
are  equal  (29.  1.)  to  two  right  angles:  but 
BAD  is  a  right  angle;  therefore  also  ADE 
is  a  right  angle;  but  the  opposite  angles 
of  parallelograms  are  equal,  (34.  1.)  there- 
fore each  of  the   opposite  angles  ABE, 

BED    is    a   right   angle;    wherefore    the      A B 

figure  ADEB  is  rectangular,  and  it  has  been  demonstrated  that  it 
is  equilateral;  it  is  therefore  a  square,  and  it  is  described  upon 
the  given  straight  line  AB.     Which  was  to  be  done. 

Cor.  Hence  every  parallelogram  that  has  one  right  angle,  has 
all  its  angles  right  angles. 

PROP.  XLVn.  THEOR. 

In  any  right  angled  triangle,  the  square  which  is 
described  upon  the  side  subtending  the  right  angle,  is 


40 


THE  ELEMENTS  OF  EUCLID. 


BOOK  I. 


equal  to  the  squares  described  upon  the  sides  which 
contain  the  right  angle. 

Let  ABC  be  a  right  angled  triangle,  having  the  right  angle 
BAG;  the  square  described  upon  the  side  BC  is  equal  to  the 
squares  described  upon  BA,  AC. 

On  BC  describe  (46.  1.)  the  square  BDEC,  and  on  BA,  AC  the 
Squares  GB,  HC:  and  through  A  draw  (31.  1.)  AL  parallel  to 
BD  or  CE,  and  join  AD,  FC:  then,  because  each  of  the  angles 
BAC,  BAG,  is  a  right   angle,  G 

(30.  def.)  the  two  straight  lines 
AC,  AG  upon  the  opposite  sides 
of  AB,  make  with  it  at  the  point 
A  the  adjacent  angles  equal  to  F 
two  right  angles:  therefore  CA 
is  in  the  same  straight  line  (14. 
1.)  with  AG:  for  the  same  rea- 
son, AB  and  AH  are  in  the  same 
straight  line;  and  because  the 
angle  DBC  is  equal  to  the  angle 
FBA,  each  of  them  being  a  right 
angle,  add  to  each  the  angle 
ABC,  and  the  whole  angle  DBA 
is  equal  (2.  Ax.)  to  the  whole 
FBC:  and  because  the  two  sides 
AB,  BD  are  equal  to  the  two  FB,  BC,  each  to  each,  and  the  angle 
DBA  equal  to  the  angle  FBC;  therefore  the  base  AD  is  equal 
(4.  1.)  to  the  base  FC,  and  the  triangle  ABD  to  the  triangle  FBC: 
now  the  parallelogram  BL  is  double  (41.  1.)  of  the  triangle  ABD, 
because  they  are  upon  the  same  base  BD,  and  between  the  same 
parallels  BD,  AL;  and  the  square  GB  is  double  of  the  triangle 
FBC,  because  these  also  are  upon  the  same  base  FB,  and  between 
the  same  parallels  FB,  GC.  But  the  doubles  of  equals  are  equal 
(6.  Ax.)  to  one  another:  therefore  the  parsi^ilelogram  BL  is  equal 
to  the  square  GB:  and  in  the  same  manner,  by  joining  AE,  BK, 
it  is  demonstrated  that  the  parallelogram  CL  is  equal  to  the 
square  HC:  therefore  the  whole  square  BDEC  is  equal  to  the 
two  squares  GB,  HC;  and  the  square  BDEC  is  described  upon 
the  straight  line  BC,  and  the  squares  GB,  HC  upon  BA,  AC: 
wherefore  the  square  upon  the  side  BC  is  equal  to  the  squares 
upon  the  sides  BA,  AC.  Therefore,  in  any  right  angled  triangle, 
&c.     Q.  E.  D. 

PROP.  XLVHL  THEOR. 

If  the  square  described  upon  one  of  the  sides  of  a 
triangle  be  equal  to  the  squares  described  upon  the 
other  two  sides  of  it;  the  angle  contained  by  these  two 
sides  is  a  right  angle. 

If  the  square  described  upon  BC,  one  of  the  sides  of  the  trian- 
gle ABC,  be  equal  to  the  squares  upon  the  other  sides  BA,  AC, 
the  angle  BAC  is  a  right  angle. 


BOOK  I.  THE  ELEMENTS  OF  EUCLID.  41 

From  the  point  A  draw  (11.  1.)  AD  at  right  angles  to  AC,  and 
make  AD  equal  to  BA,  and  join  DC:  then  because  DA  is  equal 
to  AB,  the  square  of  DA  is  equal  to  the  square  D 

of  AB:  to  each  of  these  add  the  square  of  AC: 
therefore  the  squares  of  DA,  AC  are  equal  to 
the  squares  of  13 A,  AC:  but  the  square  of  DC 
is  equal  (47.  1.)  to  the  squares  of  DA,  AC,  be- 
cause DAC  is  a  right  angle;  and  the  square  of 
BC,  by  hypothesis,  is  equal  to  the  squares  of 
BA,  AC;  therefore  the  square  of  DC  is  equal 
to  the  square  of  BC;  and  therefore  also  the 
side  DC  is  equal  to  the  side  BC.     And   be-  B  C 

cause  the  side  DA  is  equal  to  AB,  and  AC  common  to  the  two 
triangles  DAC,  BAC,  the  two  DA,  AC  are  equal  to  the  two  BA, 
AC:  and  the  base  DC  is  equal  to  the  base  BC:  therefore  the  an- 
gle DAC  is  equal  (8.  I.)  to  the  angle  BAC:  but  DAC  is  a  right 
angle;  therefore  also  BAC  is  a  right  angle.  Therefore,  if  the 
square,  &:c.  Q.  E.  D. 


THE 


ELrEMEJ^TS  OF  EUCLID. 


BOOK  II. 


BBFISTITION'S. 

I. 

EVERY  right  angled  parallelogram  is  said  to  be  contained  by 
any  two  of  the  straight  lines  which  qontain  one  of  the  right 
.  angles.  (Irr 

In  every  parallelogram,  any  of  the  parallelograms  about  a  diame- 


ter, together  with  the  two  comple 
ments  is  called  a  gnomon.     'Thus  A 
'  the  parallelogram    HG,   together 
'  with  the  complements  AF,  FC,  is 

*  the  gnomon,  which  is  more  briefly 

*  expressed  by  the  letters  AGK,  orTx 
*EHC,  which  are  at  the  opposite 

'  angles  of  the  parallelograms  which 

*  make  the  gnomon.' 


F     X  ^ 


D 


K 


B       G 


PROP.  I.  THEOR. 

If  there  be  two  straight  Hnes,  one  of  which  is  divided 
into  any  number  of  parts;  the  rectangle  contained  by 
the  two  straight  hnes,  is  equal  to  the  rectangles  con- 
tained by  the  undivided  line,  and  the  several  parts  of 
the  divided  line. 

Let  A  and  BC  be  two  straight  lines;  and  let  BC  be  divided 
into  any  parts  in  the  points  D,  E;  the  rectangle  contained  by  the 
straight  lines  A,  BC  is  equal  to  the       B  DEC 

rectangle  contained  by  A,  BD,  to- 
gether with  that  contained  by  A, 
DE,  and  that  contained  by  A,  EC. 

From  the  point  B  draw  (11.  1.) 
BF  at  right  angles  to  BC,  and  make 
BG  equal  (3.  1.)  to  A;  and  through 
G  draw  (31.  I.)  GH  parallel  to  BC; 
and  through  D,  E,  C  draw  (31.  1.) 
DK,  EL,  CH  parallel  to  BG:  then 
the  rectangle  BH  is  equal  to  the  F 
rectangles  BK,  DL,  EH;  and  BH  is  contained  by  A,  BC,  for  it  is 


BOOK  II. 


THE  ELEMENTS  OF  EUCLID, 


43 


contained  by  GB,BC,  and  GB  is  equal  to  A;  and  BK  is  contained 
by  A,  BD,  for  it  is  contained  by  GB,  BD,  of  which  GB  is  equal  to 
A;  and  I)L*is  contained  by  A,  DE,  because  DK,  that  is  (34.  1.) 
BG  is  equal  to  A;  and  in  like  manner  the  rectangle  EH  is  con- 
tained by  A,  EC:  therefore  the  rectangle  contained  by  A,  BC  is 
equal  to  the  several  rectangles  contained  by  A,  BD,  and  by  A,  DE; 
and  also  by  A,  EC.  Wherefore,  if  there  be  two  straight  lines, 
&c.  Q.  E.  D. 

.  PROP.  II.  THEOR. 

If  a  straight  line  be  divided  into  any  two  parts,  the 
rectangles  contained  by  the  whole  and  each  of  the 
parts,  are  together  equal  to  the  square  of  the  whole 
line. 

Let  the  straight  line  AB  be  divided  into 
any  two  parts  in  the  point  C;  the  rectangle 
contained  by  AB,  BC,  together  with  the  rec- 
tangle* AB,  AC,  shall  be  equal  to  the  square 
ofAB. 

Upon  AB  describe  (46.  1.)  the  square 
ADEB,  and  through  C  draw  (31.  1.)  CF,  pa- 
rallel to  AD  or  BE;  then  AE  is  equal  to  the 
rectangles  AF,  CE:  and  AE  is  the  square  of 
AB:  and  AF  is  the  rectangle  contained  by 
BA,  AC;  for  it  is  contained  by  DA,  AC,  of 
which  AD  is  equal  to  AB;  and  CE  is  contained  by  AB,  BC,  for 
BE  is  equal  to  AB;  therefore  the  rectangle  contained  by  AB,  AC, 
together  with  the  rectangle  AB,  BC,  is  equal  to  the  square  of  AB. 
If  therefore  a  straight  line,  Sec.  Q.  E.  D. 

PROP.  III.  THEOR. 
If  a  straight  line  be  divided  into  any  two  parts,  the 
rectangle  contained  by  the  whole  and  one  of  the  parts, 
is  equal  to  the  rectangle  contained  by  the  two  parts, 
together  with  the  square  of  the  foresaid  part. 

Let  the  straight  line  AB  be  divided  into  two  parts  in  the  point 
C;  the  rectangle  AB,  BC  is  equal  to  the  rectangle  AC,  CB  to- 
gether with  the  square  of  BC. 

Upon  BC  describe  (46.  1.)  the  square  AC  B 

CDEB,  and  produce  ED  to  F,  and 
through  A  draw  (31.  1.)  AF  parallel  to 
CD  or  BE;  then  the  rectangle  AE  is 
equal  to  the  rectangles  AD,  CE;  and 
'AE  is  the  rectangle  contained  by  AB, 
BC,  for  it  is  contained  by  AB,  BE,  of 
which  BE  is  equal  to  BC;  and  AD  is 
contained  by  AC,  CB,  for  CD  is  equal 
to  BC;  and  DB  is  the  square  of  BC;  p 

*  N.  B.  To  avoid  repeating  the  word  contained  too  frequently,  the  rectangle 
contained  by  two  straight  lines  AB,  AC  is  sometimes  simply  called  the  rectangle 
AB,  AC. 


I 


44 


THE  KLRMENTS  OF  EUCLID. 


BOOK  II. 


therefore  the  rectangle  AB,  BC,  is  equal  to  the  rectangle  AC,  CB, 
together  with  the  square  of  BC.  If  therefore  a  straight  line,  Sec. 
Q.  E.  D. 

PROP.  IV.  THEOR. 

If  a  straight  line  be  divided  into  any  two  parts,  the 
square  of  the  whole  line  is  equal  to  the  squares  of  the 
two  parts,  together  with  twice  the  rectangle  contained 
bj  the  parts. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  C; 
the  square  of  AB  is  equal  to  the  squares  of  AC,  CB,  and  to  twice 
the  rectangle  contained  by  AC,  CB. 

Upon  AB  describe  (46.  1.)  the  square  ADEB,  and  join  BD,  and 
through  C  draw  (31.  1.)  CGF  parallel  to  AD  or  BE  and,  through 
G  draw  HK  parallel  to  AB  or  DE:  and  because  CF  is  parallel  to 
AD,  and  BD  falls  upon  them,  the  exterior  angle  BGC  is  equal 
(29.  1.)  to  the  interior  and  opposite  angle  ADB;  but  ADB  is  equal 
(5.  1.)  to  the  angle  ABD,  because  BA  is  equal  to  AD,  being  sides 
of  a  square;  wherefore  the  angle  CGB  is      A  C  B 

equal  to  the  angle  GBC|  and  therefore  the 
side  BC  is  equal  (6.  1.)  to  the  side  CGj 
but  CB  is  equal  (34.  1.)  also  to  GK,  and 

CG  to  BK;  wherefore  the  figure  CGKB  is  Hi 7^ 1  K 

equilateral;  it  is  likewise  rectangular;  for 
CG  is  parallel  to  BK,  and  CB  meets  them; 
the  angles  KBC,  GCB  are  therefore  equal 
to  two  right  angles;  and  KBC  is  a  right 
angle;  wherefore  GCB  is  a  right  angle: 
and  therefore  also  the  angles  (34.  1.)  CGK,  GKB,  opposite  to 
these,  are  right  angles,  and  CGKB  is  rectangular:  but  it  is  also 
equilateral,  as  was  demonstrated;  wherefore  it  is  a  square,  and  it 
is  upon  the  side  CB:  for  the  same  reason  HF  also  is  a  square,  and 
it  is  upon  the  side  HG,  which  is  equal  to  AC:  therefore  HF,  CK 
are  the  squares  of  AC,  CB;  and  because  the  complement  AG  is 
equal  (43.  1.)  to  the  complement  GE,  and  that  AG  is  the  rectan- 
gle contained  by  AC,  CB,  for  GC  is  equal  to  CB;  therefore  GE  is 
also  equal  to  the  rectangle  AC,  CB;  wherefore  AG,  GE  are  equal 
to  twice  the  rectangle  AC,  CB;  and  HF,  CK  are  the  squares  of 
AC,  CB:  wherefore  the  four  figures  HF,  CK,  AG,  GE  are  equal 
to  the  squares  of  AC,  CB,  and  to  twice  the  rectangle  AC,  CB, 
but  HF,  CK;  AG,  GE  make  up  the  whole  figure  ADEB,  which 
is  the  square  of  AB:  therefore  the  square  of  AB  is  equal  to  the 
squares  of  AC,  CB,  and  twice  the  rectangle  AC,  CB.  Where- 
fore, if  a  straight  line,  &c.  Q.  E.  D. 

Cor.  From  the  demonstration  it  is  manifest,  that  the  parallel- 
ograms about  the  diameter  of  a  square  are  likewise  squares. 

PROP.  V.  THEOR. 

If  a  straight  line  be  divided  into  two  equal  parts, 
and  also  into  two  unequal  parts;  the  rectangle  con- 


■ 

G 

D 

] 

F 

E 

ROOK  II. 


THE  ELEMENTS  OF  EUCLID. 


45 


tained  by  the  unequal  parts,  together  with  the  square 
of  the  hue  between  the  points  of  section,  is  equal  to  the 
square  of  half  the  line. 

Let  the  straight  line  AB  be  dividecl  into  two  equal  parts  in  the 
point  C,  and  into  two  unequal  parts  at  the  point  D;  the  rectangle 
AD,  DB,  together  with  the  square  of  CD,  is  equal  to  the  square 
ofCB. 

Upon  CB  describe  (46.  1.)  the  square  CEFB,  join  BE,  and 
through  D  draw  (31.1.)  DUG  parallel  to  CE  or  BF;  and  through 
H  draw  KLM  parallel  to  CB  or  EF;  and  also  through  A  draw 
AK  parallel  to  CL  or  BM^  and  because  the  complement  CH  is 
equal  (43.  1.)  to  the  complement  HF,  to  each  of  these  add  DM; 
therefore  the  whole  CM  is  equal      AC  D         B 

to  the  whole  DF;  but  CM  is 
equal  (36.  1 .)  to  AL,  because  AC 
is  equal  to  CB;  therefore  also  ^ 
AL  is  equal  to  DF.  To  each  of 
these  add  CH,  and  the  whole 
AH  is  equal  to  DF  and  CH: 
but  AH  is  the  rectangle  con- 
tained by  AD,  DB,  for  DH  is 
equal  (Cor.  4.  2.)  to  DB,  and 
DF  together  with  CH  is  the  gnomon  CMG;  therefore  the  gno- 
mon CMG  is  equal  to  the  rectangle  AD,  DB:  to  each  of  these 
add  LG,  which  is  equal  (43.  1.)  to  the  square  of  CD;  therefore 
the  gnomon  CMG,  together  with  LG,  is  equal  to  the  rectangle 
AD,  DB,  together  with  the  square  of  CD:  but  the  gnomon  CMG 
and  LG  makes  up  the  whole  figure  CEFB,  which  is  the  square  of 
CB:  therefore  the  rectangle  AD,  DB,  together  with  the  square  of 
CD  is  equal  to  the  square  of  CB.  Wherefore,  if  a  straight  line, 
Sec.  Q.  E.  D. 

From  this  proposition  it  is  manifest,  that  the  difference  of  the 
squares  of  two  unequal  lines  AC,  CD,  is  equal  to  the  rectangle 
contained  by  their  sum  and  difference. 

I 

PROP.  VL  THEOR. 

If  a  Straight  line  be  bisected,  and  produced  to  any 
point;  the  rectangle  contained  by  the  whole  line  thus 
produced,  and  the  part  of  it  produced,  together  with 
the  square  of  half  the  line  bisected,  is  equal  to  the 
square  of  the  straight  line  which  is  made  up  of  the  half 
and  the  part  produced. 

Let  the  straight  line  AB  be  bisected  in  C,  and  produced  to  the 
point  D;  the  rectangle  AD,  DB,  together  with  the  square  of  CB, 
is  equal  to  the  square  of  CD. 

Upon  CD  describe  (46.  1.)  the  square  of  CEFD,  join  DE,  and 
through  B  draw  (3M.)  BHG  parallel  to  CE  or  DF,  and  through 


46 


THE  ELEMENTS  OF  EUCLID. 


BOOK  II. 


H 

/ 

/ 

H  draw  KLM  parallel  to  AD  or  EF,  and  also  through  A  draw 

AK  parallel  to  CL  or  DM:  and      A C  B     D 

because  AC  is  equal  to  CB,  the      . 
rectangle  AL  is  equal  (43.  1.)  to      i  i 

CH^  but  CH  is  equal  (36.  1.)  to  K  

HF;  therefore  also  AL  is  equal 
to  HF:  to  each  of  these  add  CM; 
therefore  the  whole  AM  is  equal 
to  the  gnomon  CMG:  and  DM  E  G      F 

is  the  rectangle  contained  by  AD,  DB,  for  DM  is  equal  (Cor.  4. 
2.)  to  DB:  therefore  the  gnomon  CMG  is  equal  to  the  rectangle 
AD,  DB,  add  to  each  of  these  LG,  which  is  equal  to  the  square 
of  CB,  therefore  the  rectangle  AD,  DB,  together  with  the  square 
of  CB,  is  equal  to  the  gnomon  CMG  and  the  figure  LG:  but  the 
gnomon  CMG  and  LG  make  up  the  whole  figure  CEFD,  which 
is  the  square  of  CD;  therefore  the  rectangle  AD,  DB,  together 
with  the  square  of  CB,  is  equal  to  the  square  of  CD.  Wherefore, 
if  a  straight  line,  8cc.   Q.  E.  D. 


PROP.  Vn.  THEOR. 

If  a  straight  line  be  divided  into  any  two  parts,  the 
squares  of  the  whole  line,  and  of  one  of  the  parts  are 
equal  to  twice  the  rectangle  contained  by  the  whole 
and  that  part,  together  with  the  square  of  the  other 
part. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  the 
point  C;  the  squares  of  AB,  BC  are  equal  to  twice  the  rectangle 
AB,  BC,  together  with  the  square  of  AC. 

Upon  AB  describe  (46.  1.)  the  square  ADEB,  and  construct 
the  figure  as  in  the  preceding  propositions:  and  because  AG  is 
equal  (43.  1.)  to  GE,  add  to  each  of  them  CK;  the  whole  AK  is 
therefore  equal  to  the  whole  CE;  there-       A  C  B 

fore  AK,  CE  are  double  of  AK:  but  AK, 
CE  are  the  gnomon  AKF,  together  with 
the  square  CK;  therefore  the  gnomon 
AKF,  together  with  the  square  CK,  is  1'^ 
double  of  AK:  but  twice  the  rectangle 
AB,  BC  is  double  of  AK,  for  BK  is  equal 
(Cor.  4.  2.)  to  BC:  therefore  the  gnomon 
AKF,  together  with  the  square  CK,  is 
equal  to  twice  the  rectangle  AB,  BC:  to      D  F  E 

each  of  these  equals  add  HF,  which  is  equal  to  the  square  of  AC: 
therefore  the  gnomon  AKF,  together  with  the  squares  CK,  HF, 
is  equal  to  twice  the  rectangle  AB,  BC,  and  the  square  of  AC: 
but  the  gnomon  AKF,  together  with  the  squares  CK,  HF,  make 
up  the  whole  figure  ADEB  and  CK,  which  are  the  squares  of  AB 
and  BC:  therefore  the  squares  of  AB  and  BC  are  equal  to  twice 
the  rectangle  AB,  BC,  together  with  the  square  AC.  Wherefore, 
if  a  straight  line,  Sec.  Q.  E.  D. 


BOOK  II. 


THE  ELEMENTS  OF  EUCLID. 


47 


PROP.  VIII.    THEOR. 

If  a  straight  line  be  divided  into  any  two  parts,  four 
times  the  rectangle  contained  by  the  whole  line,  and 
one  of  the  parts,  together  with  the  square  of  the  other 
part,  is  equal  to  the  square  of  the  straight  line  which 
is  made  up  of  the  whole  and  that  part. 

Let  the  straight  line  AB  be  divided  into  any  two  parts  in  the 
point  C;  four  times  the  rectangle  AB,  BC,  together  with  the 
square  of  AC,  is  equal  to  the  square  of  the  straight  line  made  up 
of  AB  and  BC  together. 

Produce  AB  to  D,  so  that  BD  be  equal  to  CB,  and  upon  AD 
describe  the  square  AEFD;  and  construct  two  figures  such  as  in 
the  preceding.  Because  CB  is  equal  to  BD,  and  thatCB  is  equal 
(34.  1.)  to  GK,  and  BD,  to  KN;  therefore  GK  is  equal  to  KN; 
for  the  same  reason,  PR  is  equal  to  RO;  and  because  CB  is  equal 
to  BD,  and  GK  to  KN,  the  rectangle  CK  is  equal  (36.  1.)  to  BN, 
and  GR  to  RN;  but  CK  is  equal  (43.  1.)  to  RN,  because  they  are 
the  complements  of  the  parallelogram  CO;  therefore  also  BN  is 
eq\aal  to  GR;  and  the  four  rectangles  BN,  CK,  GR,  RN  are 
therefore  equal  to  one  another,  and  so  are  quadruple  of  one  of 
them  CK:  again,  because  CB  is  equal  to  BD,  and  that  BD  is 
equal  (Cor.  4.  2.)  to  BK,that  is,  to  CG;  C     B 

and  CB  equal  to  GK,  that  (Cor.  4.  2.)  A 
is,  to  GP;  therefore  CG  is  equal  to  GP:       j  G 

and  because  CG  is  equal  to  GP,  and  PR  M  r 
to  RO,  the  rectangle  AG  is  equal  to 
MP,  and  PL  to  RF:  but  MP  is  equal  X 
(43.  1.)  to  PL,  because  they  are  the 
complements  of  the  parallelogram  ML; 
wherefore  AG  is  equal  also  to  RF: 
therefore  the  four  rectangles  AG,  MP, 
PL,  RF  are  equal  to  one  another,  and  so 
are  quadruple  of  one  of  them  AG.  And  it  was  demonstrated 
that  the  four  CK,  BN,  GR,  and  RN  are  quadruple  of  CK:  there- 
fore the  eight  rectangles  which  contain  the  gnomon  AOH  are 
quadruple  of  AK:  and  because  AK  is  the  rectangle  contained  by 
AB,  BC,  for  BK  is  equal  to  BC,  four  times  the  rectangle  AB,  BC 
is  quadruple  of  AK:  but  the  gnomon  AOH  was  demonstrated  to 
be  quadruple  of  AK:  therefore  four  times  the  rectangle  AB,  BC 
is  equal  to  the  gnomon  AOH.  To  each  of  these  add  XH,  which 
is  equal  (Cor.  4.  2.)  to  the  square  of  AC:  therefore  four  times  the 
rectangle  AB,  BC,  together  with  the  square  of  AC,  is  equal  to 
the  gnomon  AOH  and  the  square  XH:  but  the  gnomon  AOH 
and  XH  make  up  the  figure  AEFD,  which  is  the  square  of  AD: 
therefore  four  times  the  rectangle  AB,  BC,  together  with  the 
square  of  AC  is  equal  to  the  square  of  AD,  that  is,  of  AB  and 
BC  added  together  in  one  straight  line.  Wherefore,  if  a  straight 
line,  Sec.  Q.  E.  D. 


48  THE  ELEMENTS  OF  EUCLID.  BOOK  II. 

PROP.  IX.  THEOR. 

If  a  straight  line  be  divided  into  two  equal,  and  also 
into  two  unequal  parts;  the  squares  of  the  two  unequal 
parts  are  together  double  of  the  square  of  half  the  line, 
and  of  the  square  of  the  line  between  the  points  of  sec- 
tion. 

Let  the  straight  line  AB  be  divided  at  the  point  C  into  two 
equal,  and  at  D  into  two  unequal  parts:  the  squares  of  AD,  DB 
are  together  double  of  the  squares  of  AC,  CD. 

From  the  point  C  draw  (11.  1.)  CE  at  right  angles  to  AB,  and 
make  it  equal  to  AC  or  CB,  and  join  EA,  EB;  through  D  draw 
(31.  1.)  DF  parallel  to  CE,  and  through  F  draw  FG  parallel  to 
AB;  and  join  AF:  then,  because  AC  is  equal  to  CE,  the  angle 
EAC  is  equal  (5.  1.)  to  the  angle  AEC;  and  because  the  angle 
ACE  is  a  right  angle,  the  two  others,  AEC,  EAC  together  make 
one  right  angle  (32.  1.);  and  they  are  equal  to  one  another;  each 
of  them  therefore  is  half  of  a  right  an-  E 

gle.  For  the  same  reason  each  of  the 
angles  CEB,  EBC  is  half  a  right  angle; 
and  therefore  the  whole  AEB  is  a  right  X  Q 

angle:  and  because  the  angle  GEF  is 
half  a  right  angle,  and  EGF  a  right 
angle,  for  it  is  equal  (29.  1.)  to  the  in- 
terior and  opposite  angle  ECB,  the  re-  A  CD  B 
maining  angle  EFG  is  half  a  right  angle;  therefore  the  angle 
GEF  is'^equal  to  the  angle  EFG,  and  the  side  EG  equal  (6.  1.)  to 
the  side  GF :  again,  because  the  angle  at  B  is  half  a  right  angle, 
and  FDB  half  a  right  angle,  for  it  is  equal  (29.  1,)  to  the  interior 
and  opposite  angle  ECB,  the  remaining  angle  BED  is  half  a  right 
angle;  therefore  the  angle  at  B  is  equal  to  the  angle  BED,  and  the 
side  DF  to  (6.  1.)  the  side  DB:  and  because  AC  is  equal  to  CE, 
the  square  of  AC  is  equal  to  the  square  of  CE;  therefore  the 
squares  of  AC,  CE   are  double  of  the  square  of  AC:  but  the 
square  of  EA  is  equal  (47.  1.)  to  the  squares  of  AC,  CE,  because 
ACE  is  a  right  angle;  therefore  the  square  of  EA  is  double  of 
the  square  of  AC:  again,  because  EG  is  equal  to  GF,  the  square 
of  EG  is  equal  to  the  square  of  GF;  therefore  the  squares  of  FG, 
GF  are  double  of  the  square  of  GF;  but  the  square  of  EF  is  equal 
to  the  squares  of  EG,  GF;  therefore  the  square  of  EF  is  double 
of  the  square  of  GF;  and  GF  is  equal  (34.  1.)  to  CD;  therefore 
the  square  of  EF  is  double  of  the  square  of  CD:  but  the  square 
of  AE  is  likewise  double  of  the   square  of  AC:   therefore  the 
squares  of  AE,  EF  are  double  of  the  squares  of  AC,  CD:  and  the 
square  of  AF  is  equal  (47.  1.)  to  the  squares  of  AE,  EF,  because 
AEF  is  a  right  angle;  therefore  the  square  of  AF  is  double  of  the 
squares  of  AC,  CD:  but  the  squares  of  AD,  DF  are  equal  to  the 
square  of  AF,  because  the  angle  ADF  is  a  right  angle;  therefore 
the  squares  of  AD,  DF  are  double  of  the  squares  of  AC,  CD:  and 
DF  is  equal  to  DB;  therefore  the  squares  of  AD,  DA  are  double 
of  the  squares  of  AC,  CD.     If  therefore  a  straight  line,  &c.     Q, 
E.  D. 


BOOK  II.  THE  ELEMENTS  OF  EUCLID.  49 


PROP.  X.  THEOR. 

If  a  straight  line  be  bisected,  and  produced  to  any 
point,  the  square  of  the  whole  line  thus  produced  and 
the  square  of  the  part  of  it  produced,  are  together  dou- 
ble of  the  square  of  half  the  line  bisected,  and  of  the 
square  of  the  line  made  up  of  the  half  and  the  part 
produced. 

Let  the  straight  line  AB  be  bisected  in  C,  and  produced  to  the 
point  D;  the  squares  of  AD,  DB  are  double  of  the  squares  of 
AC,  CD. 

From  the  point  C  draw  (11.  1.)  CE  at  rif^ht  angles  to  AB:  and 
make  it  equal  to  AC  or  CB,  and  join  AE,  EB;  through  E  draw 
(31.  1.)  EF  parallel  to  AB,  and  through  D  draw  DF,  parallel  to 
CE:  and  because  the  straight  line  EF  meets  the  parallels  EC,FD, 
the  angles  CEF,  EFD  are  equal  (29.  1.)  to  two  right  angles^  and 
therefore  the  angles  BEF,  EFD  are  less  than  two  right  angles: 
but  straight  lines  which  with  another  straight  line  make  the  inte- 
rior angles  upon  the  same  side  less  than  two  right  angles  do  meet 
(12.  Ax.)  if  produced  far  enough:  therefore  EB,  FD  shall  met't, 
if  produced  towards  B,  D:  let  them  meet  in  G,  and  join  AG: 
then,  because  AC  is  equal  to  CE,  the  angle  CEA  is  equal  (5.  1.) 
to  the  angle  EAC:  and  the  angle  ACE  is  a  right  angle:  therefore 
each  of  the  angles  CEA,  EAC  is  half  a  right  angle  (32.  1.):  for 
the  same  reason,  each  of  the  angles  CEB,  EBC  is  half  a  right  an- 
gle: therefore  AEB  is  a  right  angle:  and  because  EBC  is  half  a 
right  angle,  DBG  is  also  (15.  1.)  half  a  right  angle,  for  they  are 
vertically  opposite;  but  BDG  is  a  right  angle,  because  it  is  equal 
(29.  1.)  to  the  alternate  angle  DCE;  therefore  the  remaining  an- 
gle DGB  is  half  a  right  angle,  and  is  therefore  equal  to  the  angle 
DBG;  wherefore  also  the  side  BD  is  equal  (6.  1.)  to  the  side  DG: 
again,  because  EGF  is  half  a  right  angle,  and  that  the  angle  at  F 
is  a  right  angle,  because  it  is  E  F 

equal  (34.  1.)  to  the  opposite 
angle  ECD,  the  remaining  an- 
gle FEG  is  half  a  right  angle, 
and  equal  to  the  angle  EGF; 
wherefore  also  the  side  GF  is 
equal  (6.  1.)  to  the  side  FE.  A 
And  because  EC  is  equal  to 
CA,  the  square  of  EC  is  equal 
to  the  square  of  CA;  therefore 
the  squares  of  EC,  CA  are  dou- 
ble of  the  square  of  CA:  but  the  square  of  EA  is  equal  (47.  1 ,)  to 
the  squares  of  EC,  CA;  therefore  the  square  of  EA  is  double  of  the 
square  of  AC:  again,  because  GF  is  equal  to  FE,  the  square  of 
GF  is  equal  to  the  square  of  FE:  and  therefore  the  squares  of 
GF,  FE  are  double  of  the  square  of  EF:  but  the  square  of  EG  is 
equal  (47.  1.)  to  the  squares  of  GF,  FE;  therefore  the  square  of 
G 


50 


THii  ELEMENTS  OF  EUCLID. 


BOOK  11 


EG  is  double  of  the  square  of  EF:  and  EF  is  equal  to  CD5 
wherefore  the  square  of  EG  is  double  of  the  square  of  CD:  but 
it  was  demonstrated,  that  the  square  of  EA  is  double  of  the 
square  of  AC;  therefore  the  squares  of  AE,  EG  are  double  of 
the  squares  of  AC,  CD:  and  the  square  of  AG  is  equal  (47.  1.) 
to  the  squares  of  AE,  EG;  therefore  the  square  of  AG  is  double 
of  the  squares  of  AC,  CD:  but  the  squares  of  AD,  GD  are  equal 
(47.  1.)  to  the  square  of  AG;  therefore  the  squares  of  AD,  DG 
are  double  of  the  squares  of  AC,  CD:  but  DG  is  equal  to  DB5 
therefore  the  squares  of  AD,  DB  are  double  of  the  squares  of 
AC,  CD.     Wherefore,  if  a  straight  line.  Sec.  Q.  E.  D. 


PROP.  XL  PROB. 

To  divide  a  given  straight  line  into  two  parts,  so  that 
the  rectangle  contained  by  the  whole  and  one  of  the 
parts  shall  be  equal  to  the  square  of  the  other  part. 

Let  AB  be  the  given  straight  line:  it  is  required  to  divide  it 
into -two  parts,  so  that  the  rectangle  contained  by  the  whole  and 
one  of  the  parts  shall  be  equal  to  the  square  of  the  other  part. 

Upon  AB  describe  (46.  1.)  the  square  ABDC;  bisect  (10.  1.) 
AC  in  E,  and  join  BE;  produce  CA  to  F,  and  make  (3.  1.)  EF 
equal  to  EB;  and  upon  AF  describe  (46.  1.)  the  square  FGHAj 
AB  is  divided  in  H,  so  that  the  rectangle  AB,  BH  is  equal  to  the 
square  of  Ali. 

Produce  GPI  to  K;  because  the  straight  line  AC  is  bisected 
in  E,  and  produced  to  the  point  F,  the  rectangle  CF,  FA,  together 
with  the  square  of  AE,  is  equal  (6.  2.)  to  the  square  of  EF:  but 
EF  is  equal  to  EB;  therefore  the  rectangle  CF,  FA,  together 
with  the  square  of  AE,  is  equal  to  the  square  of  EB:  and  the 
squares  of  BA,  AE  are  equal  (47.  1 .)  to  the      F  G 

square  of  EB,  because  the  angle  EAB  is  a 
right  angle;  therefore  the  rectangle  CF,  FA, 
together  with  the  square  of  AE,  is  equal  to 
the  squares  of  BA,  AE:  take  away  the 
square  of  AE,  which  is  common  to  both, 
therefore  the  remaining  rectangle  CF,  FA  A 
is  equal  to  the  square  of  AB;  and  the  figure 
FK  is  the  rectangle  contained  by  CF,  FA, 
for  AF  is  equal  to  FG;  and  AD  is  the 
square  of  AB;  therefore  FK  is  equal  to  AD:  E 
take  away  the  common  part  AK,  and  the  re- 
mainder FH  is  equal  to  the  remainder  HI), 
and  FID  is  the  rectangle  contained  by  AB, 
BH,  for  AB  is  equal  to  BD;  and  FH  is  the      C  K         D 

square  of  AH:  therefore  the  rectangle  AB,  BH  is  equal  to  the 
square  of  AH:  wherefore  the  straight  line  AB  is  divided  in  H,  so 
that  the  rectangle  AB,  BH  is  equal  to  the  S(iuare  of  AH.  Which 
was  to  be  done. 


l?OOK  IT. 


THE    ELEMENTS  OF    EUCLID.  51 


PROP.  XII.  THEOR. 


In  obtuse  angled  triangles,  if  a  perpendicular  be 
drawn  from  any  of  the  acute  angles  to  the  opposite 
side  produced,  the  square  of  the  side  subtending  the 
obtuse  angle  is  greater  than  the  squares  of  the  sides 
containing  the  obtuse  angle,  by  twice  the  rectangle 
contained  by  the  side  upon  which,  when  produced,  the 
perpendicular  falls,  and  the  straight  line  intercepted 
without  the  triangle  between  the  perpendicular  and  the 
obtuse  angle. 

Let  ABC  be  an  obtuse  angled  triangle,  having  the  obtuse  an- 
gle ACB,  and  from  the  point  A  let  AD  be  drawn  (12.  1.)  perpen- 
diculur  to  BC  produced:  the  square  of  AB  is  greater  than  the 
squares  of  AC,  CB  by  twice  the  rectangle  BC,  CD. 

Because  the  straight  line  BD  is  divided  into  two  parts  in  the 
point  C,  the  square  of  BD  is  equal 

(4.  2.)  to  the  squares  of  BC,  CD,  ^  A 

and  twice  the  rectangle  BC,  CD: 
to  each  of  these  equals  add  the 
square  of  DA;  and  the  squares  of 
BD,  DA  are  equal  to  the  squares 
of  BCx,  CD,  DA,  and  twice  the  rect- 
angle BC,  CD:  but  the  square  of 
BA  is  equal  (47.  1.)  to  the  squares 
of  BD,  DA,  because  the  angle  at  D 
is  a  right  angle;  and  the  square  of 
CA  is  equal  (47.  1.)  to  the  squares  of  CD,  DA:  therefore  the 
square  of  BA  is  equal  to  the  squares  of  BC,  CA,  and  twice  the 
rectangle  BC,  CD;  that  is,  the  square  of  BA  is  greater  than  the 
squares  of  BC,  CA,  by  twice  the  rectangle  BC,  CD.  Therefore, 
in  obtuse  angled  triangles,  8cc.  Q.  E.  D. 

PROP.  XIII.  THEOR. 

In  every  triangle,  the  square  of  the  side  subtending 
any  of  the  acute  angles  is  less  than  the  squares  of  the 
sides  containing  that  angle,  by  twice  the  rectangle  con- 
tained by  either  of  these  sides,  and  the  straight  line 
intercepted  between  the  perpendicular  let  fall  upon  it 
from  the  opposite  angle,  and  the  acute  angle.* 

Let  ABC  be  any  triangle,  and  the  angle  at  B  one  of  its  acute 
angles,  and  upon  BC  one  of  the  sides  containing  it,  let  fall  the 
perpendicular  (12.  1.)  AD  from  the  opposite  angle:  the  square 
of  AC,  opposite  to  the  angle  B,  is  less  than  the  squares  of  CB, 
BA,  by  twice  the  rectangle  CB,  BD. 

*  See  Note. 


52 


THE  ELEMENTS  OF  EUCLID. 


BOOK  II, 


First,  Let  AD  fall  within  the  triangle  ABC;  and  because  the 
straight  line  CJ^  is  divided  into  two  parts 
in  the  point  Dy'lhe  squares  of  CB,  BD  are 
equal  (7.  2.)  to  twice  the  rectangle  contain- 
ed by  CB,  BD,  and  the  square  of  DC:  to 
each  of  these  equals  add  the  square  of 
AD;  therefore  the  squares  of  CB,  BD,  DA 
are  equal  to  twice  the  rectangle  CB,  BD, 
and  the  squares  of  AD,  DC:  but  the  square 
of  AB  is  equal  (47.  1.)  to  the  squares  of 
BD,  DA,  because  the  angle  BD  A  is  a  right  B 
angle,  and  the  square  of  AC  is  equal  to  the  squares  of  AD,  DC: 
therefore  the  squares  of  CB,  BA  are  equal  to  the  square  of  AC, 
and  twice  the  rectangle  CB,  BD,  that  is,  the  square  of  AC  alone 
is  less  than  the  squares  of  CI3,  BA  by  twice  the  rectangle  CB,  BD. 

Secondly,  Let  AD  fall  without  the  tri-  A 

angle  ABC:  then,  because  the  angle  at 
D  is  a  right  angle,  the  angle  ACB  is 
greater  (16.  I.)  than  a  right  angle;  and 
therefore  the  square  of  AB  is  equal 
(12.  2.)  to  the  squares  of  AC,  CB,  and 
twice  the  rectangle  BC,  CD:  to  these 
equals  add  the  square  of  BC,  and  the 
squares  of  AB,  BC  are  equal  to  the 
square  of  AC,  and  twice  the  square  of  g 
BC,  and  twice  the  rectangle  BC,  CD: 
but  because  BD  is  divided  into  two  parts  in  C,  the  rectangle  DB, 
BC  is  equal  (3.  2.)  to  the  rectangle  BC,  CD  and  the  square  of  BC: 
and  the  doubles  of  these  are  equal:  therefore  the  squares  of  AB, 
BC  are  equal  to  the  square  of  AC,  and  twice  the  rectangle  DB, 
BC:  therefore  the  square  of  AC  alone  is  less  than  the  squares  of 
AB,  BC  by  twice  the  rectangle  DB,  BC.  A 

Lastly,  Let  the  side  AC  be  perpendicular  to  BC; 
then  is  BC  the  straight  line  between  the  perpendicular 
and  the  acute  angle  at  B;  and  it  is  manifest  that  the 
square  of  AB,  BC  are  equal  (47.  I.)  to  the  square  of 
AC  and  twice  the  square  of  BC.  Therefore,  in  every 
triangle,  &c.  Q.  E.  D. 


PROP.  XIV.    PROB. 

To  describe  a  square  that  shall  be  equal  to  a  given 
rectilineal  figure."* 

Let  A  be  the  given  rectilineal  figure;  it  is  required  to  describe 
a  square  that  shall  be  equal  to  A. 

iSescribe  (45.  1.)  the  rectangular  parallelogram  BCDE  equal 
to  the  rectilineal  figure  A.     If  then  the  sides  of  it  BE,  ED  are 


*  See  Note. 


BOOK  II. 


THE  ELEMENTS  OF  EUCLID. 


equal  to  one  another, 
it  is  a  square,  and 
what  was  required 
is  now  done:  but  if 
they  are  not  equal, 
produce  one  of  them 
BE  to  F,  and  make 
EF  equal  to  ED,  and 
bisect  BF  in  G;  and 
from  the  centre  G, 
at  the  distance  GB,  or  GF,  describe  the  semicircle  BHF,  and 
produce  DE  to  H,  and  join  GH;  therefore,  because  the  straight 
line  BF  is  divided  into  two  equal  parts  in  the  point  G,  and  into 
two  unequal  at  E,  the  rectangle  BE,  EF,  together  with  the  square 
of  EG,  is  equal  (5.  2.)  to  the  square  of  GF:  but  GF  is  equal  to 
GH;  therefore  the  rectangle  BE,  EF,  together  with  the  square  of 
EG,  is  equal  to  the  square  of  GH;  but  the  squares  of  HE,  EG 
are  equal  (47.  1.)  to  the  square  of  GH;  therefore  the  rectangle 
BE,  EF  together  with  the  square  of  EG,  is  equal  to  the  squares 
of  HE,  EG:  take  away  the  square  of  EG,  which  is  common  to 
both;  and  the  remaining  rectangle  BE,  EF  is  equal  to  the  square 
of  EH:  but  the  rectangle  contained  by  BE,  EF  is  the  parallelo- 
gram BD,  because  EF  is  equal  to  ED;  therefore  BD  is  equal  to 
the  square  of  EH;  but  BD  is  equal  to  the  rectilineal  figure  A; 
therefore  the  rectilineal  figure  A  is  equal  to  the  square  of  EH: 
wherefore  a  square  has  been  made  equal  to  the  given  rectilineal 
figure  A,  viz.  the  square  described  upon  EH.  Which  was  to  be 
done. 


THE 


ELEMENTS  OF  EUCLID. 


BOOK  III. 


DEFIITITIOITS. 


I. 

EquAL  circles  are  those  of  which  the  diameters   are  equal,  or 
from  the  centres  of  which  the  straight  lines  to  the  circumfe- 
rences are  equal. 
'  This  is  not  a  definition,  but  a  theorem,  the  truth  of  which  is 

evident;  for,  if  the  circles  be  applied  to  one  another,  so  that  their 

centres  coincide,  the  circles  must  likewise   coincide,  since  the 

straight  lines  from  the  centre  are  equal.' 

II. 

A  straight  line  is  said  to  touch  a 
circle,  when  it  meets  the  circle, 
and   being   produced    does    not 


cut  It. 


III. 


Circles  are  said  to  touch  one  ano- 
ther, which  meet,  but  do  not  cut 
one  another. 

IV. 

Straight  lines  are  said  to  be  equally  distant 
from  the  centre  of  a  circle,  when  the  per- 
pendiculars drawn  to  them  from  the  cen- 
tre are  equal. 

V. 

And  the  straight  line  on  which  the  greater 
perpendicular  falls,  is  said  to  be  farther 
from  the  centre. 

VI. 

A  segment  of  a  circle  is  the  figure  con- 
tained by  a  straight  line  and  the  cir- 
cumference it  cuts  oflD 


BOOK  III. 


THE  ELEMEN'IS  OF  EUCLID. 


55 


VII. 

"The  angle  of  a   segment  is  that  which  is  contained  by  the 
straight  line  and  the  circumference." 

VIII. 

An  angle  in  a  segment  is  the  angle  con- 
tained by  two  straight  lines  drawn  from 
any  point  in  the  circumference  of  the 
segment,  to  the  extremities  of  the 
straight  line  which  is  the  base  of  the 
segment. 

And-  an  angle  is  said  to  insist  or  stand 
upon  the  circumference  intercepted  be- 
tween tlic  straight  lines  that  contain 
the  angle. 

X. 

The  sector  of  a  circle  is  the  figure  con- 
tained by  two  straight  lines  drawn  from 
the  centre,  and  the  circumference  be- 
tween them. 


XI. 

Similar  segments  of  a  circle,  arc 
those  in  which  the  angles  are 
equal,  or  which  contain  equal 
angles. 


PROP.  I.    PROB. 


# 


To  find  the  centre  of  a  given  circle.* 

Let  ABC  be  the  given  circle;  it  is  required  to  find  its  centre. 

Draw  within  it  any  straight  line  AB,  and  bisect  (10.  1.)  it  in  D; 
from  the  point  D  draw  (U.  1.)  DC  at  right  angles  to  AB,  and 
produce  it  to  E,  and  bisect  CE  in  F:  the  point  F  is  the  centre 
of  the  circle  ABC. 

For,  if  it  be  not,  let,  if  possible,  G  be  the  centre,  and  join  GA, 
GD,  GB:  then,  because  DA  is  equal  to  DB,  and  DG  common  to 
the  two  triangles  ADG,  BDG,  the  two  sides  AD,  DG  are  e(jual 
to  the  two  BD,  DG,  each  to  each;  and  the  base  GA  is  equal  to 
the  base  GB,  because  they  are  drawn  from  the  centre  Gif  there- 
fore the  angle  ADG  is  equal  (8.  1.)  to  the  angle  GDB:  but  when 
a  straight  line  standing  upon  another  straight  line  makes  the  ad,- 


*  See  Note. 

t  N.  B.  Whenever  the  expression  "straight  lines  from  the  centre,"  or 
"  drawn  from  the  centre,"  occurs,  it  is  to  be  understood  liiat  they  are  drawn  to 
the  circumference. 


56 


THE  ELEMENTS  OF  EUCLID. 


BOOK  111. 


jacent  angles  equal  to  one  another,  each  of 
the  angles  is  a  right  angle:  (10.  def.  1.) 
therefore  the  angle  GDB  is  a  right  angle: 
but  FDB  is  likewise  a  right  angle;  where- 
fore the  angle  FDB  is  equal  to  the  angle 
GDB,  the  greater  to  the  less,  which  is  im- 
possible: therefore  G  is  not  the  centre  of 
the  circle  ABC:  in  the  same  manner  it 
can  be  shown,  that  no  other  point  but  F  is 
the  centre;  that  is,  F  is  the  centre  of  the 
circle  ABC.     Which  was  to  be  found. 

Cor.  From  this  it  is  manifest,  that  if  in 
a  circle  a  straight  line  bisect  another  at  right  angles,  the  centre 
of  the  circle  is  in  the  line  which  bisects  the  other. 


E 


PROP.  II.  THEOR. 

If  any  two  points  be  taken  in  the  circumference  of 
a  circle,  the  straight  hne  which  joins  them  shall  fall 
within  the  circle. 

Let  ABC  be  a  circle,  and  A,  B  any  two  points  in  the  circum- 
ference; the  straight  line  drawn  from  A  to  C 
B  shall  fall  within  the  circle. 

For,  if  it  do  not,  let  it  fall,  if  possible, 
without,  as  AEB;  find  (1.  3.)  D  the  centre 
of  the  circle  ABC,  and  join  AD,  DB,  and 
produce  DF,  any  straight  line  meeting  the 
circumference  AB,  to  E:  then  because  DA 
is  equal  to  DB,  the  angle  DAB  is  equal 
(5.  1.)  to  the  angle  DBA;  and  because  AE, 
a  side  of  the  triangle  DAE,  is  produced  to 
B,  the  angle  DEB  is  greater  (16.  1.)  than 
the  angle  DAE:  but  DAE  is  equal  to  the  angle  DBE:  therefore 
the  angle  DEB  is  greater  than  the  angle  DBE:  but  to  the  greater 
angle  the  greater  side  is  opposite  (19.  1.);  DB  is  therefore  greater 
than  DE:  but  DB  is  equal  to  DF;  wherefore  DF  is  greater  than 
DE,  the  less  than  the  greater,  which  is  impossible:  therefore  the 
straight  line  drawn  from  A  to  B  does  not  fall  without  the  circle. 
In  the  same  manner  it  may  be  demonstrated  that  it  does  not  fall 
upon  the  circumference;  it  falls  tlierefore  within  it.  Wherefore, 
if  any  two  points.  Sec.  Q.  E.  D, 


PROP.  IIL  THEOR. 


If  a  straight  line  drawn  through  the  centre  of  a  cir- 
cle bisect  a  straight  line  in  it  which  does  not  pass 
through  the  centre,  it  shall  cut  it  at  right  angles,  and, 
if  it  cuts  it  at  right  angles,  it  shall  bisect  it. 

Let  ABC  be  a  circle;  and  let  CD,  a  straight  line  drawn  through 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


57 


the  centre,  bisect  any  straight  line  AB,  which  does  not  pass 
through  the  centre,  in  the  point  F:  it  cuts  it  also  at  right  angles. 

Take  (1.3.)  E  the  centre  of  the  circle,  and  join  EA,  EB.  Then, 
because  AF  is  equal  to  FB,  and  FE  common  to  the  two  triangles 
AFE,  BFE,  there  are  two  sides  in  the  one  equal  to  two  sides  in 
the  other,  and  the  base  EA  is  equal  to  the 
base  EB:  therefore  the  angle  AFE  is  equal 
(8.  1.)  to  the  angle  BFE:  but  when  a 
straight  line  standing  upon  another  makes 
the  adjacent  angles  equal  to  one  another, 
each  of  them  is  a  right  (lO.  def.  1.)  angle:  | 
therefore  each  of  the  angles  AFE,  BFE  is  a 
right  angle;  wherefore  the  straight  line  CD, 
drawn  through  the  centre  bisecting  another  a^ 
AB  that  does  not  pass  through  the  centre, 
cuts  the  same  at  right  angles. 

But  let  CD  cut  AB  at  right  angles:  CD  also  bisects  it,  that  is, 
AF  is  equal  to  FB. 

The  same  construction  being  made,  because  EA,  EB  from  the 
centre  are  equal  to  one  another,  the  angle  EAF  is  equal  (5.  1.)  to 
the  angle  EBF:  and  the  right  angle  AFE  is  equal  to  the  right 
angle  BFE:  therefore,  in  the  two  triangles,  EAF,  EBF,  there  are 
two  angles  in  one  equal  to  two  angles  in  the  other,  and  the  side 
EF,  which  is  opposite  to  one  of  the  equal  angles,  in  each,  is  com- 
mon to  both;  therefore  the  other  sides  are  equal  (26.  1.):  AF 
therefore  is  equal  to  FB.  Wherefore,  if  a  straight  line.  Sec.  Q. 
E.D. 

PROP.  IV.  THEOR. 

If  in  a  circle  two  straight  lines  cut  one  another 
which  do  not  both  pass  through  the  centre,  they  do  not 
bisect  each  other. 

Let  ABCD  be  a  circle,  and  AC,  BD  two  straight  lines  in  it 
which  cut  one  another  in  the  point  E,  and  do  not  both  pass 
through  the  centre:  AC,  BD  do  not  bisect  one  another. 

For,  if  it  is  possible,  let  AE  be  equal  to  EC,  and  BE  to  ED;  if 
one  of  the  lines  pass  through  the  centre,  it  is  plain  that  it  cannot 
be  bisected  by  the  other  which  does  not 
pass  through  the  centre;  but,  if  neither 
of  them  pass  through  the  centre,  take 
(l.  3.)  F  the  centre  of  the  circle,  and 
join  EF:  and  because  FE,  a  straight 
line  through  the  centre,  bisects  another  A  \ 
AC  which  does  not  pass  through  the 
centre,  it  shall  cut  it  at  right  (3.  3.)  an- 
gles; w;herefore  FEA  is  a  right  angle:  B  ^ — "^      C 

again,  because  the  straight  line  FE  bi- 
sects the  straight  line  BD  which  does  not  pass  through  the  cen- 
tre, it  shall  cut  it  at  right  (3.  3.)  angles;  wherefore  FEB  is  a  right 
angle,  and  FEA  was  shown  to  be  a  right  angle;  therefore  FEA 
H 


/ 


( 


58 


THE  ELEMENTS  OF  EUCLID. 


BOOK  III. 


is  equal  to  the  angle  FEB,  the  less  to  the  greater,  which  is  impos- 
sible: therefore  AC,  BD  do  not  bisect  one  another.  Wherefore, 
if  in  a  circle,  Sec.  Q.  E.  D. 

PROP.  V.  THEOR. 

If  two  circles  cut  one  another,  they  shall  not  have 
the  same  centre. 

Let  the  two  circles  ABC,  CDG  cut  one  another  in  the  points 
B,  C;  they  have  not  the  same  centre. 

For,  if  it  be  possible,  let  E  be  their  centre:  join  EC,  and  draw 
any  straight  line  EFG  meeting  them  C 

in  F  and  G;  and  because  E  is  the 
centre  of  the  circle  ABC,  CE  is  equal 
to  EF:  again,  because  E  is  the  centre 
of  the  circle  CDG,  CE  is  equal  to 
EG:  but  CE  was  shown  to  be  equal 
to  EF^  therefore  EF  is  equal  to  EG,  A 
the  less  to  the  greater,  which  is  im- 
possible: therefore  E  is  not  the  centre 
of  the  circles  ABC,  CDG.  Where- 
fore, if  two  circles.  Sec.  Q.  E.  D. 


G 


PROP.  VI.  THEOR. 


If  two  circles  touch  one  another  internally,  they 
shall  not  have  the  same  centre. 


Let  the  two  circles  ABC,  CDE  touch  one  another  inter 
the  point  C:  they  have  not  the  same  centre. 

For,  if  they  can,  let  it  be  F;  join  FC,  and  draw  any  strai 
FEB  meeting  them  in  E  and  B^  and  C 

because  F  is  the  centre  of  the  circle 
ABC,  CF  is  equal  to  FB;  also,  be- 
cause F  is  the  centre  of  the  circle 
CDE,  CF  is  equal  to  FB:  and  CF 
was  shown  equal  to  FB;  therefore 
FE  is  equal  to  FB,  the  less  to  the 
greater,  which  is  impossible:  where- 
fore F  is  not  the  centre  of  the  circles 
ABC,  CDE.  Therefore,  if  two  cir- 
cles, &c.  Q.  E.  D. 


nally  in 
ght  line 


B 


PROP.  Vn.  THEOR. 


If  any  point  be  taken  in  the  diameter  of  a  circle, 
which  is  not  the  centre,  of  all  the  straight  lines  which 
can  be  drawn  from  it  to  the  circumference,  the  great- 
est is  that  in  which  the  centre  is,  and  the  other  part  of 
that  diameter  is  the  least;  and,  of  any  others,  that 


BOOK  III.  THE  ELEMENTS  OF  EUCLID.  59 

which  is  nearer  to  the  hne  which  passes  through  the 
centre  is  always  greater  than  one  more  remote;  and 
from  the  same  point  there  can  be  drawn  only  two 
straight  lines  that  are  equal  to  one  another,  one  upon 
each  side  of  the  shortest  line. 

Let  ABCD  be  a  circle,  and  AD  its  diameter,  in  which  let  any 
point  F  be  taken  which  is  not  the  centre;  let  the  centre  be  E;  of 
all  the  straight  lines  FB,  FC,  FG,  Sec.  that  can  be  drawn  from  F 
to  the  circumference,  FA  is  the  greatest,  and  FD,  the  other  part 
of  the  diameter  AD,  is  the  least:  and  of  the  other,  FB  is  greater 
than  FC,  and  FC  than  FG. 

Join  BE,  CE,  GE:  and  because  two  sides  of  a  triangle  are 
greater  (20.  1.)  than  the  third,  BE,  EF  are  greater  than  B¥^  but 
AE  is  equal  to   EB;    therefore  AE,  EF,  A 

that  is,  AF,  is  greater  than  BF:  again,  be- 
cause BE  is  equal  to  CE,  and  FE  common 
to  the  triangles  BEF,  CEF,  the  two  sides  C^ 
BE,  EF  are  equal  to  the  two  CE,  EF;  but 
the  angle  BEF  is  greater  than  the  angle 
CEF;  therefore  the  base  BF  is  greater 
(24.  1.)  than  the  base  FC:  for  the  same 
reason,  CF  is  greater  than  GF:  again,  be- 
cause GF,  FE  are  greater  (20.  1.)  than  EG, 
and  EG  is  equal  to  ED;  GF,  FE  are 
greater  than  ED:  take  away  the  common 
part  FE,  and  the  remainder  GF  is  greater  than  the  remainder 
FD:  therefore  FA  is  the  greatest,  and  FD  the  least  of  all  the 
straight  lines  from  F  to  the  circumference;  and  BF  is  greater 
than  CF,  and  CF  than  GF. 

Also  there  can  be  drawn  only  tw^o  equal  straight  lines  from  the 
point  F  to  the  circumference,  one  upon  each  side  of  the  shortest 
line  FD:  at  the  point  E,  in  the  straight  line  EF,  make  (23.  1.)  the 
angle  FEH  equal  to  the  angle  GEF,  and  join  FH:  then  because 
GE  is  equal  to  EH,  and  EF  common  to  the  two  triangles  GEF, 
HEF;  the  two  sides  GE,  EF  are  equal  to  the  two  HE,  EF;  and 
the  angle  GEF  is  equal  to  the  angle  HEF;  therefore  the  base  FG 
is  equal  (4.  1.)  to  the  base  FH:  but,  besides  FH,  no  other  straight 
line  can  be  drawn  from  F  to  the  circumference  equal  to  FG,  for, 
if  there  can,  let  it  be  FK;  and  because  FK  is  equal  to  FG,  and  FG 
to  FH,  FK  is  equal  to  FH:  that  is,  a  line  nearer  to  that  which 
passes  through  the  centre,  is  equal  to  one  which  is  more  remote: 
which  is  impossible.  Therefore,  if  any  point  be  taken,  &c.  Q. 
E.  D. 

PllOP.  VHI.  THEOR. 

If  any  point  be  taken  without  a  circle,  and  straight 
lines  be  drawn  from  it  to  the  circumference,  whereof 
one  passes  through  the  centre,  of  those  which  fall  upon 
the  concave  circumference,  the  greatest  is  that  which 


60 


THE  ELEMENTS  OF  EUCLID. 


BOOK  III. 


passes  through  the  centre,  and,  of  the  rest,  that  which 
is  nearer  to  that  through  the  centre  is  always  greater 
than  the  more  remote:  but  of  those  which  fall  upon 
the  convex  circumference,  the  least  is  that  between  the 
point  without  the  circle  and  the  diameter;  and,  of  the 
rest,  that  which  is  nearer  to  the  least  is  always  less 
than  the  more  remote:  a;id  only, two  equal  straight 
lines  can  be  drawn  from  the  point  into  the  circumfe- 
rence, one  upon  each  side  of  the  least. 

Let  ABC  be  a  circle,  and  D  any  point  without  it,  from  which 
let  the  straight  lines  DA,  DE,  DF,  DC  be  drawn  to  the  circum- 
ference, whereof  DA  passes  through  the  centre.  Of  those  which 
fall  upon  the  concave  part  of  the  circumference  AEFC,  the  great- 
est is  AD,  which  passes  through  the  centre;  and  the  nearer  to  it 
is  always  greater  than  the  more  remote,  viz.  DE  than  DF,  and 
DF  than  DC;  but  of  those  which  fall  upon  the  convex  circum- 
ference HLKG,  the  least  is  DG  between  the  point  D  and  the  dia- 
meter AG;  and  the  nearer  to  it  is  always  less  than  the  more  re- 
mote, viz.  DK  than  DL,  and  DL  than  DH. 

Take  (l.  3.)  M  the  centre  of  the  circle  ABC,  and  join  ME,  MF, 
MC,  MK,  ML,  MH:  and  because  AM  is  equal  to  ME,  add  MD 
to  each,  therefore  AD  is  equal  to  EM,  MD ;  but  EM,  MD  are 
greater  (20.  1.)  than  ED ;  therefore  also  AD  is  greater  than  ED : 
again,  because  ME  is  equal  to  MF,  and  MD  common  t©  the  tri- 
angles EMD,  FMD  :  EM,  MD  are  D 
equal  to  FM,  MD  ;  but  the  angle 
EMD  is  greater  than  the  angle 
FMD;  therefore  the  base  ED  is 
greater  (24.  1.)  than  the  base  FD : 
in  like  manner  it  may  be  shown  that 
FD  is  greater  than  CD:  therefore 
DA  is  the  greatest :  and  DE  great- 
er than  DF,  and  DF  than  DC:  and 
because  MK,  KD  are  greater  (20.  1.) 
than  MD,  and  MK  is  equal  to  MG, 
the  remainder  KD  is  greater  (4.  Ax.) 
than  the  remainder  GD,  that  is,  GD 
is  less  than  KD :  and  because  MK, 
DK,  are  drawn  to  the  point  K  with- 
in the  triangle  MLD,  from  M,  D, 
the  extremities  of  its  side  MD ; 
MK,  KD  are  less,  (21.  1.)  than 
ML,  LD  virhereof  MK  is  equal  to  E  A 
ML;  therefore  the  remainder  DK  is  less  than  the  remainder  DL: 
in  like  manner  it  may  be  shown  that  DL  is  less  than  DH ;  there- 
fore DG  is  the  least,  and  DK  less  than  DL,  and  DL  than  DH; 
also  there  can  be  drawn  only  two  equal  straight  lines  from  the 
point  D  to  the  circumference,  one  upon  each  side  of  the  least ;  at 
the  point  M,  in  the  straight  line  MD,  make  the  angle  DMB  equal 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


61 


G 


to  the  angle  DMK,  and  join  DB ;  and  because  MK  is  equal  to  MB, 
and  MD  common  to  the  triangles  KMD,  BMD,  the  two  sides 
KM,  MD  are  equal  to  the  two  BM,  MD ;  and  the  angle  KMD  is 
equal  to  the  angle  BMD ;  therefore  the  base  DKis  equal  to  (4.  1.) 
the  base  DB  :  but,  besides  DB,  there  can  be  no  straight  line  drawn 
from  D  to  the  circumference  equal  to  DK  :  for,  if  there  can,  let  it 
be  DN ;  and  because  DK  is  equal  to  DN,  and  also  to  DB;  there- 
fore DB  is  equal  to  DN,  that  is,  the  nearer  to  the  least  equal  to 
the  more  remote,  which  is  impossible.  If,  therefore,  any  point, 
&c.  Q.  E.  D. 

PROP.  IX.  TIIEOR. 

If  a  point  be  taken  within  a  circle,  from  which  there 
fall  more  than  two  equal  straight  lines  to  the  circumfer- 
ence, that  point  is  the  centre  of  the  circle. 

Let  the  point  D  be  taken  within  the  circle  ABC,  from  which 
to  the  circumference  there  fall  more  than  two  equal  straight  lines, 
viz.  DA,  DB,  DC :  the  point  D  is  the  centre  of  the  circle. 

For,  if  not,  let  E  be  the  centre,  join 
DE,  and  produce  it  to  the  circumfer- 
ence in  F,  G;  then  FG  is  a  diameter 
of  the  circle  ABC :  and  because  in 
FG,  the  diameter  of  the  circle  ABC, 
there  is  taken  the  point  D  which  is  not 
the  centre,  DG  shall  be  the  greatest 
line  from  it  to  the  circumference,  and 
DC  greater  (7.  3.)  than  DB,  and  DB 
than  DA;  but  they  are  likewise  equal,  A  B 

which  is  impossible :  therefore  E  is  not  the  centre  of  the  circle 
ABC  :  in  like  manner  it  may  be  demonstrated,  that  no  other  point 
but  D  is  the  centre;  D  therefore  is  the  centre.  Wherefore,  if  a 
point  be  taken,  Sec.  Q.  E.  D. 

PROP.  X.  THEOR. 

One  circumference  of  a  circle  cannot  cut  another  in 
more  than  two  points. 

If  it  be  possible,  let  the  circumfer- 
ence FAB  cut  the  circumference 
DEF  in  more  than  two  points,  viz,  in 
B,  G,  F;  take  the  centre  K  of  the  cir- 
cle ABC,  and  join  KB,  KG,  KF ;  and 
because  within  the  circle  DEF  there 
is  taken  the  point  K,  from  which  to 
the  circumference  DEF  fall  more 
than  two  equal  straight  lines  KB, 
KG,  KF,  the  point  K  is  (9.  3.)  the 
centre  of  the  circle  DEF  :  but  K  is  C 

also  the  centre  of  the  circle  ABC  ; 
therefore  the  barne  point  is  the  centre  of  two  circles  that  cut  one 


62 


THE  ELEMENTS  OF  EUCLID. 


BOOK  III. 


another,  which  is  impossible  (5.  3.)  Therefore  one  circumference 
of  a  circle  cannot  cut  another  in  more  than  two  points.     Q.  E.  D. 


PROP.  XI.  THEOR. 

If  two  circles  touch  each  other  internally,  the 
straight  line  which  joins  their  centres  being  produced 
shall  pass  through  the  point  of  contact. 

Let  the  two  circles  ABC,  ADE  touch  each  other  internally  in 
the  point  A,  and  let  F  be  the  centre  of  the  circle  ABC,  and  G  the 
centre  of  the  circle  ADE:  the  straight  line  A 

which  joins  the  centres  F,  G,  being  pro- 
duced, passes  through  the  point  A. 

For,  if  not,  let  it  fall  otherwise,  if  possi- 
ble, as  FGDH,  and  join  AF,  AG:  and  be- 
cause AG,  GF  are  greater  (20.  1.)  than 
FA,  that  is  than  FH,  for  FA  is  equal  to 
FH,  both  being  from  the  same  centre^  take 
away  the  common  part  FG;  therefore  the 
remainder  AG  is  greater  than  the  remain- 
der GH:  but  AG  is  equal  to  GD;  there- 
fore GD  is  greater  than  GH,  the  less  than  the  greater,  which  is 
impossible.  Therefore  the  straight  line  which  joins  the  points 
F,  G  cannot  fall  otherwise  than  upon  the  point  A,  that  is,  it  must 
pass  through  it.     Therefore,  if  two  circles,  &c.  Q.  E.  D. 


B 


PROP.  XII.  THEOR. 

If  two  circles  touch  each  other  externally,  the 
straight  line  which  joins  their  centres  shall  pass  through 
the  point  of  contact. 

Let  the  two  circles  ABC,  ADE  touch  each  other  externally 
in  the  point  A;  and  let  F  be  the  centre  of  the  circle  ABC,  and 
G  the  centre  of  ADE:  the  straight  line  which  joins  the  points  F, 
G  shall  pass  through  the  point  of  contact  A. 

For,  if  not,  let  it  pass  otherwise,  if  possible,  as  FCDG,  and 
join  FA,  AG:  and  because  F  is  the  centre  of  the  circle  ABC,  AF 
is  equal  to  FC:  also  because 
G  is  the  centre  of  the  circle 
ADE,  AG  is  equal  to  GD:   , 
therefore  FA,  AG  are  equal  / 
to  FC,  DG;  wherefore  the , 
whole  FG  is    greater   than  \ 
FA,  AG:  but  it  is  also  less  \ 
(20.  1.);  which   is   impossi-     \ 
ble:    therefore    the   straight 
line  which  joins  the  points  F,  G  shall  not  pass  otherwise  than 
through  the  point  of  contact  A,  that  is,  it  must  pass  through  it. 
Therefore,  if  two  circles.  Sec.  Q.  E.  D. 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


PROP.  XIII.  THEOR. 

One  circle  cannot  touch  another  in  more  points  than 
one,  whether  it  touches  it  on  the  inside  or  outside.* 

For,  if  it  be  possible,  let  the  circle  EBF  touch  the  circle  ABC 
in  more  points  than  one,  and  first  on  the  inside,  in  the  points  B,  D; 
join  BD,  and  draw  (10.  11.  I.)  GH  bisecting  BD  at  right  angles. 
Therefore,  because  the  points  B,  D  are  in  the  circumference  of 

H 


B 


each  of  the  circles,  the  straight  line  BD  falls  within  each  (^2.  3.) 
of  them:  and  their  centres  are  (Cor.  1.  3.)  in  the  straight  line  GH 
which  bisects  BD  at  right  angles;  therefore  GH  passes  through 
the  point  of  contact  (11.  3.);  but  it  does  not  pass  through  it,  be- 
cause the  points  B,  D  are  without  the  straight  line  GH,  which  is 
absurd:  therefore  one  circle  cannot  touch  another  on  the  inside  in 
more  points  than  one. 

Nor  can  two  circles  touch  one  another  on  the  outside  in  more 
than  one  point:  for,  if  it  be  possible,  let  the  circle  ACK  touch  the 
circle  ABC  in  the  points  A,  C,  and  join  AC:  therefore,  because 
the  two  points  A,  C  are  in  the  circumference 
of  the  circle  ACK,  the  straight  line  AC 
which  joins  them  shall  fall  within  (2.  3.) 
the  circle  ACK:  and  the  circle  ACK  is 
without  the  circle  ABC;  and  therefore  the 
straight  line  AC  is  without  this  last  circle; 
but  because  the  points  A,  C  are  in  the  cir- 
cumference of  the  circle  ABC,  the  straight 
line  AC  must  be  within  (2.  3.)  the  same  cir- 
cle, which  is  absurd;  therefore  one  circle 
cannot  touch  another  on  the  outside  in  more 
than  one  point:  and  it  has  been  shown  that 
they  cannot  touch  on  the  inside  in  more 
points  than  one.  Therefore  one  circle.  Sec. 
Q.  E.  D. 

PROP.  XIV.  THEOR. 

EquAL  straight  hues  in  a  circle  are  equally  distant 
from  the  centre;  and  those  which  are  equally  distant 
from  the  centre  are  equal  to  one  another. 


*  See  Note. 


64  THE  ELEMENTS  OF  EUCLID.  BOOK  III. 

Let  the  straight  lines  AB,  CD,  in  the  circle  ABDC,  be  equal 
to  one  another:  they  are  equally  distant  from  the  centre. 

Take  E  the  centre  of  the  circle  ABDC,  anci  from  it  draw  EF, 
EG  perpendiculars  to  AB,  CD^  then,  because  the  straight  line 

EF,  passing  through  the  centre,  cuts  the  straight  line  AB,  which 
does  not  pass  through  the  centre,  at  right  C 
angles,  it  also  bisects  (3.  3.)  it:  wherefore 
AF  is  equal  to  FB,  and  AB  double  of  AF. 
For  the  same  reason,  CD  is  double  of  CG; 
and  AB  is  equal  to  CD;  therefore  AF  is 
equal  to  CG:  and  because  AE  is  equal  to 
EC,  the  square  of  AE  is  equal  to  the 
square  of  EC;  but  the  squares  of  AF,  FE 
are  equal  (47.  1.)  to  the  square  of  AE,  be- 
cause the  angle  AFE  is  a  right  angle; 
and,  for  the  like  reason,  the  squares  of 

EG,  GC  are  equal  to  the  square  of  EC:  therefore  the  squares  of 
AF,  FE  are  equal  to  the  squares  of  CG,  GE,  of  which  the  square 
of  AF  is  equal  to  the  square  of  CG,  because  AF  is  equal  to  CG; 
therefore  the  remaining  square  of  FE  is  equal  to  the  remaining 
square  of  EG,  and  the  straight  line  EF  is  therefore  equal  to  EG: 
but  straight  lines  in  a  circle  are  said  to  be  equally  distant  from 
the  centre,  when  the  perpendiculars  drawn  to  them  from  the  cen- 
tre are  equal  (4.  def.  3.)  therefore  ABCD  are  equally  distant  from 
the  centre. 

Next,  if  the  straight  lines  AB,  CD  be  equally  distant  from  the 
centre;  that  is,  if  FE  be  equal  to  EG,  AB  is  equal  to  CD:  for,  the 
same  construction  being  made,  it  may,  as  before,  be  demonstrated, 
that  AB  is  double  of  AF,  and  CD  double  of  CG,  and  that  the 
squares  of  EF,  FA  are  equal  to  the  squares  of  EG,  GC;  of  which 
the  square  of  FE  is  equal  to  the  square  of  EG,  because  FE  is 
equal  to  EG;  therefore  the  remaining  square  of  AF  is  equal  to 
the  remaining  square  of  CG;  and  the  straight  line  AF  is  there- 
fore equal  to  CG:  and  AB  is  double  of  AF,  and  CD  double  of 
CG;  wherefore  AB  is  equal  to  CD.  Therefore  equal  straight 
lines,  See.  Q.  E.  D. 


PROP.  XV.  THEOR. 

The  diameter  is  the  greatest  straight  Hne  in  a  circle; 
and,  of  all  others,  that  which  is  nearer  to  the  centre  is 
always  greater  than  one  more  remote;  and  the  greater 
is  nearer  to  the  centre  than  the  less.'^ 

•^  See  Note. 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


65 


Let  ABCD  be  a  circle,  of  which  the  dia-  A  B 

meter  is  AD,  and  the  centre  E;  and  let  BC  , 

be  nearer  to  the  centre  than   FG;  AD    is     ^ / 
greater  than  any  straight  line  BC  which  is     /  N. 
not  a  diameter,  and  BC  greater  than  FG. 

From  the  centre  draw  EH,  EK  perpendi- 
cular to  BC,  FG,  and  join  EB,  EC,  EFj  and 
because  AE  is  equal  to  EB,  and  ED  to  EC, 
AD  is  equal  to  EB,  EC;  but  EB,  EC  are 
greater  (20.  1.)  than  BC;  wherefore  also 
AD  is  greater  than  BC. 

And,  because  BC  is  nearer  to  the  centre  than  FG,  EH  is  less 
(5.  def.  3.)  than  EK ;  but,  as  it  was  demonstrated  in  the  prece- 
ding, BC  is  double  of  BH,  and  FG  double  of  FK,  and  the  squares 
of  EH,  HB  are  equal  to  the  squares  of  EK,  KF,  of  which  the 
square  of  EH  is  less  than  the  square  of  EK,  because  EH  is  less 
than  EK;  therefore  the  square  of  BH  is  greater  than  the  square 
of  FK,  and  the  straight  line  BH  greater  than  FK ;  and  therefore 
BC  is  greater  than  FG. 

Next,  let  BC  be  greater  than  FG;  BC  is  nearer  to  the  centre 
than  FG,  that  is,  the  same  construction  being  made,  EH  is  less 
than  EK :  because  BC  is  greater  than  FG,  BH  likewise  is  great- 
er than  KF :  and  the  squares  of  BH,  HE  are  equal  to  the  squares 
of  FK,  KE,  of  which  the  square  of  BH  is  greater  than  the  square 
of  FK,  because  BH  is  greater  than  FK;  therefore  the  square  of 
EH  is  less  than  the  square  of  EK,  and  the  straight  line  EH  less 
than  EK.     Wherefore  the  diameter,  kc.  Q.  E.  D. 


PROP.  XVI.  THEOR. 

The  straight  line  drawn  at  right  angles  to  the  diam- 
eter of  a  circle,  from  the  extremity  of  it,  falls  without 
the  circle;  and  no  straight  line  can  be  drawn  between 
that  straight  line  and  the  circumference  from  the  ex- 
tremity, so  as  not  to  cut  the  circle;  or  which  is  the 
same  thing,  no  straight  line  can  make  so  great  an 
acute  angle  with  the  diameter  at  its  extremity,  or  so 
small  an  angle  with  the  straight  line  which  is  at  right 
angles  to  it  as  not  to  cut  the  circle."^ 

Let  ABC  be  a  circle,  the  centre  of  which  is  D,  and  the  diame- 
ter AB;  the  straight  line  drawn  at  right  angles  to  AB  from  its 
extremity  A,  shall  fall  without  the  circle. 


See  Note. 


66 


THE  ELEMENTS  OF  EUCLID. 


BOOK  III 


For,  if  it  does  not,  let  it  fall,  if  pos- 
sible, within  the  circle,  as  AC,  and 
draw  DC  to  the  point  C  where  it  meets 
the  circunnference  ;  and  because  DA 
is  equal  to  DC,  the  angle  DAC  is 
equal  (5.  1.)  to  the  angle  ACD  ;  but  B 
DAC  is  a  right  angle,  therefore  ACD  - 
is  a  right  angle,  and  the  angles  DAC, 
ACD  are  therefore  equal  to  two  right 
angles;  which  is  impossible  (17.  1.): 
therefore  the  straight  line  drawn  from  A  at  right  angles  to  BA 
does  not  fall  within  the  circle;  in  the  same  manner,  it  may  be 
demonstrated,  that  it  does  not  fall  upon  the  circumference  ;  there- 
fore it  must  fall  without  the  circle,  as  AE. 

And  between  the  straight  line  AE  and  the  circumference  no 
straight  line  can  be  drawn  from  the  point  A  which  does  not  cut 
the  circle:  for,  if  possible,  let  FA  be  between  them,  and  from  the 
point  D  draw  (12.  1.)  DG  perpendicular  to  FA,  and  let  it  meet 
the  circumference  in  H  :  and  because  AGD  is  a  right  angle,  and 
EAG  less  (19.  1.)  than  a  right  angle :  DA  is  greater  (19.  I.)  than 
DG:  but  DA  is  equal  to  DH  :  there-  E 

fore  DH  is  greater  than  DG,  the  less 
than  the  greater,  which  is  impossible: 
therefore  no  straight  line  can  be 
drawn  from  the  point  A  between 
AE  and  the  circumference,  which  does 
not  cut  the  circle;  or,  which  amounts 
to  the  same  thing,  however  great  an 
acute  angle  a  straight  line  makes  with 
the  diameter  at  the  point  A,  or  how- 
ever small  an  angle  it  makes  with  AE, 
the  circumference  passes  between  that 
straight  line  and  the  perpendicular 
AE.  'And  this  is  all  that  is  to  be 
understood,  when,  in  the  Greek  text  and  translations  from  it,  the 
angle  of  the  semicircle  is  said  to  be  greater  than  any  acute  rec- 
tilineal angle,  and  the  remaining  angle  less  than  any  rectilineal 
angle.* 

CoR.  From  this  it  is  manifest,  that  the  straight  line  which  is 
drawn  at  right  angles  to  the  diameter  of  a  circle  from  the  ex- 
tremity of  it,  touches  the  circle ;  and  that  it  touches  it  only  in  one 
point,  because,  if  it  did  meet  the  circle  in  two,  it  would  fall  with- 
in it  (2.  3.).  'Also  it  is  evident  that  there  can  be  but  one  straight 
line  which  touches  the  circle  in  the  same  point.' 

PROP.  XVII.  PROB. 

To  draw  a  straight  line  from  a  given  point,  either  with- 
out or  in  the  circumference,  which  shall  toucli  a  given 
circle. 

First,  let  A  be  a  given  point  without  the  given  circle  BCD: 
it  is  required  to  draw  a  straight  line  from  A  which  shall  touch 
the  circle. 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


67 


Find  (1.  3.)  the  centre  E  of  the  circle,  and  join  AE;  and  from 
the  centre  E,  at  the  distance  EA,  describe  the  circle  AFG;  from 
the  point  D  draw  (11.  1.)  DF  at  right  angles  to  EA,  and  join  EBF, 
AB.     AB  touches  the  circle  BCD. 

Because  E  is  the  centre  of  the 
circles  BCD,  AFG,  EA  is  equal  to 
EF;  and  ED  to  EB;  therefore  the 
two  sides  AE,  EB  are  equal  to  the 
two  FE,  ED,  and  they  contain  the 
angle  at  E  common  to  the  two  tri-  G 
angles  AEB,  FED;  therefore  the 
base  DF  is  equal  to  the  base  AB, 
and  the  triangle  FED  to  the  tri- 
angle AEB,  and  the  other  angles 
to  the  other  angles  (4.  1.);  there- 
fore the  angle  EBx\,  is  equal  to  the  angle  EDF:  but  EDF  is  a 
right  angle,  wherefore  EBA  is  a  right  angle;  and  EB  is  drawn 
from  the  centre;  but  a  straight  line  drawn  from  the  extremity  of 
a  diameter,  at  right  angles  to  it,  touches  the  circle  (Cor.  16.  3.): 
therefore  AB  touches  the  circle;  and  it  is  drawn  from  the  given 
point  A.    Which  was  to  be  done. 

But,  if  the  given  point  be  in  the  circumference  of  the  circle,  as 
the  point  D,  draw  DE  to  the  centre  E,  and  EF  at  right  angles  to 
DE;  DF  touches  the  circle  (Cor.  16  3.). 


PROP.  XVIII.  THEOR. 

If  a  straight  line  touch  a  circle,  the  straight  line 
drawn  from  the  centre  to  the  point  of  contact,  shall  be 
perpendicular  to  the  line  touching  the  circle. 

Let  the  straight  line  DE  touch  the  circle  ABC  in  the  point  C; 
take  the  centre  F,  and  draw  the  straight  line  FC;  FC  is  perpendi- 
cular to  DE. 

For,  if  it  be  not,  from  the  point  F  draw  FBG  perpendicular  to 
DE;  and  because  FGC  is  a  right  angle,  GCF  is  (17.  1.)  an  acute 
angle;  and  to  the  greater  angle  the  greatest  (19.  1.)  side  is  oppo- 
site; therefore  FC  is  greater  than  FG 
but  FC  is  equal  to  FB;  therefore  FB  is 
greater  than  FG,  the  less  than  the  great- 
er, which  is  impossible;  wherefore  FG 
is  not  perpendicular  to  DE:  in  the  same 
manner  it  may  be  shown,  that  no  other 
is  perpendicular  to  it  besides  FC,  that 
is,  FC  is  perpendicular  to  DE.  There- 
fore, if  a  straight  line,  &c.    Q.  E.  D. 


68 


THE    ELEMENTS  OF  EUCLID. 


BOOK  III. 


PROP.  XIX.  THEOR. 


If  a  straight  line  touch  a  circle,  and  from  the  point 
of  contact  a  straight  line  be  drawn  at  right  angles  to 
the  touching  line,  the  centre  of  the  circle  shall  be  in 
that  line. 


if  possible;  and  join  CF:  be- 
A 


Let  the  straight  line  DE  touch  the  circle  ABC  in  C,  and  from 
C  let  CA  be  drawn  at  right  angles  to  DE;  the  centre  of  the  circle 
is  in  CA. 

For,  if  not,  let  F  be  the  centre 
cause  DE  touches  the  circle  ABC, 
and  FC  is  drawn  from  the  centre 
to  the  point  of  contact,  FC  is  per- 
pendicular (18.  3.)  to  DE;  there- 
fore FCE  is  a  right  angle;  but  ACE 
is  also  a  right  angle;  therefore  the 
angle  FCE  is  equal  to  the  angle 
ACE,  the  less  to  the  greater,  which 
is  impossible:  wherefore  F  is  not 
the  centre  of  the  circle  ABC;  in 
the  same  manner  it  may  be  shown, 
that  no  other  poirft  which  is  not  in 
CA,  is  the  centre;  that  is,  the  centre  is  in  CA 
straight  line,  &c.    Q.  E.  D. 


D 


C  E 

Therefore  if  a 


PROP.  XX.  THEOR. 

The  angle  at  the  centre  of  a  circle  is  double  of  the 
angle  at  the  circumference,  upon  the  same  base,  that  is, 
upon  the  same  part  of  the  circumference."^ 

Let  ABC  be  a  circle,  and  BEC  an  angle  at  the  centre,  and  BAG 
an  angle  at  the  circumference,  which  have  the  same  circumfe- 
rence BC  for  their  base;  the  angle  BEC  is 
double  of  the  angle  BAC. 

First,  let  E  the  centre  of  the  circle  be 
within  the  angle  BAC,  and  join  AE,  and 
produce  it  to  F;  because  EA  is  equal  to 
EB,  the  angle  EAB  is  equal  (5.  1.)  to  the 
angle  EBA;  therefore  the  angles  EAB, 
EBA  are  double  of  the  angle  EAB,  but  the 
angle  BEF  is  equal  (32.  1.)  to  the  angles 
EAB,  EBA;  therefore  also  the  angle  BEF 
is  double  of  the  angle  EAB:  for  the  same 
reason,  the  angle  FEC  is  double  of  the  angle  EAC:  therefore  the 
whole  angle  BEC  is  double  of  the  whole  angle  BAC. 


See  Note. 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


(i9 


Again;  let  E  the  centre  of  the  circle 
be  without  the  angle  BDC,  and  join  DE 
and  produce  it  to  G.  It  may  be  demon- 
strated, as  in  the  first  case,  that  the  angle 
GEC  is  double  of  the  angle  GDC,  and 
that  GEB  a  part  of  the  first  is  double  of 
GDB  a  part  of  the  other;  therefore  the 
remaining  angle  BEC  is  double  of  the 
remaining  angle  BDC.  Therefore  the 
angle  at  the  centre,  Sec.  Q.  E.  D. 


A 


PROP.  XXI.  THEOR. 

The  angles  in  the  same  segment  of  a  circle  are  equal 
to  one  another ."^ 

Let  ABCD   be   a  circle,   and   BAD,  A  E 

BED  angles  in  the  same  segment 
BAED:  the  angles  BAD,  BED  are 
equal  to  one  another. 

Take  F  the  centre  of  the  circle  ABCD; 
and,  first,  let  the  segment  BAED  be 
greater  than  a  semicircle,  and  join  BF, 
FD:  and  because  the  angle  BFD  is  at  the 
centre,  and  the  angle  BAD  at  the  cir- 
cumference, and  that  they  have  the  same 
part  of  the  circumference,  viz.  BCD,  for 
their  base;  therefore  the  angle  BFD  is  double  (20.  3.)  of  the  angle 
BAD:  for  the  same  reason,  the  angle  BFD  is  double  of  the  angle 
BED:  therefore  the  angle  BAD  is  equal  to  the  angle  BED. 

But,  if  the  segment  BAED  be  not  greater  than  a  semicircle,  let 
BAD,  BED  be  angles  in  it;  these  also  ~ 

are  equal  to  one  another:  draw  AF  to 
the  centre,  and  produce  it  to  C,  and  join 
CE:  therefore  the  segment  BADC  is 
greater  than  a  semicircle;  and  the  an- 
gles in  it,  BAC,  BEC  are  equal,  by  the 
first  case;  for  the  same  reason  because 
CBED  is  greater  than  a  semicircle,  the 
angles  CAD,  CED  are  equal:  therefore 
the  whole  angle  BAD  is  equal  to  the 
whole  angle  BED.  Wherefore  the  an- 
gles in  the  same  segment.  Sec.  Q.  E.  D. 


PROP.  XXII.  THEOR. 


The    opposite    angles  of  any  quadrilateral  figure 
described  in  a  circle,  are  together  equal  to  two  right 


angles. 


*  See  Note. 


70  THE  ELEMENTS  OF  EUCLID.  BOOK  III. 

Let  ABCD  be  a  quadrilateral  figure  in  the  circle  ABCD;  any 
two  of  its  opposite  angles  are  together  equal  to  two  right  angles. 

Join  AC,  BD;  and  because  the  three  angles  of  every  triangle 
are  equal  (32.  1.)  to  two  right  angles,  the  three  angles  of  the  tri- 
angle CAB,  viz.  the  angles  CAB,  ABC,  BCA  are  equal  to  two 
right  angles:  but  the  angle  CAB  is  equal  D 

(21.  3.)  to  the  angle  CDB,  because  they 
are  in  the  same  segment  BADC,  and  the 
angle  ACB  is  equal  to  the  angle  ADB, 
because  they  are  in  the  same  segment 
ADCB:  therefore  the  whole  angle  ADC 
is  equal  to  the  angles  CAB,  ACB:  to 
each  of  these  equals  add  the  angle  ABC: 
therefore  the  angles  ABC,  CAB,  BCA 
are  equal  to  the  angles  ABC,  ADC:  but 
ABC,  CAB,  BCA  are  equal  to  two  right  angles;  therefore  also 
the  angles  ABC,  ADC  are  equal  to  two  right  angles;  in  the  same 
manner,  the  angles  BAD,  DCB  may  be  shown  to  be  equal  to  two 
right  angles.     Therefore  the  opposite  angles,  &c.  Q.  E.  D. 

PROP.  XXIII.  THEOR. 

Upon  the  same  straight  hne,  and  upon  the  same  side 
of  it,  there  cannot  be  two  similar  segments  of  circles, 
not  coinciding  with  one  another."^ 

If  it  be  possible,  let  the  two  similar  segments  of  circles,  viz. 
ACB,  ABD  be  upon  the  same  side  of  the  same  straight  line  AB, 
not  coinciding  with  one  another:  then,  because  the  circle  ACB 
cuts  the  circle  ADB  in  the  two  points  A,  B,  D 

they  cannot  cut  one  another  in  any  other 
point  (lO.  3.):  one  of  the  segments  must 
therefore  fall  within  the  other;  let  ACB  fall 

within   ADB,   and    draw   the   straight   line 

BCD,  and  join  CA,  DA:   and  because  the   A  B 

segment  ACB  is  siniilar  to  the  segment  ADB,  and  that  similar 
segments  of  circles  contain  (11.  def.  3.)  equal  angles;  the  angle 
ACB  is  equal  to  the  angle  ADB,  the  exterior  to  the  interior, 
which  is  impossible  (16.  1.).  Therefore,  there  cannot  be  two  si- 
milar segments  of  a  circle  upon  the  same  side  of  the  same  line, 
which  do  not  coincide.  Q.  E.  D. 

PROP.  XXIV.  THEOR. 

Similar  segments  of  circles  upon  equal  straight  lines, 
are  equal  to  one  another."^ 

Let  AEB,  CFD  be  similar  segments  of  circles  upon  the  equal 
straight  lines  AB,  CD:  the  segment  AEB  is  equal  ro  the  segment 

CFD.  ; 

Sec  Noteti. 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


71 


For,  if  the  segment 
AEB  be  applied  to  the 
segment  CFD,  so  as  the 
point  A  be  on  C,  and 

the    straight    line   AB  A  B       C 

upon  CD,  the  point  B  shall  coincide  with  the  point  D,  because 
AB  is  equal  to  CD:  therefore  the  straight  line  AB  coinciding 
with  CD,  the  segment  AEB  must  (23.  3.)  coincide  with  the  seg- 
ment CFD,  and  therefore  is  equal  to  it.  Wherefore  similar  seg- 
ments, &c.  Q..  E.  D. 


PROP.  XXV.    PROB. 

A  SEGMENT  of  a  circle  being  given  to  describe  the 
circle  of  which  it  is  the  segment."^ 

Let  ABC  be  the  given  segment  of  a  circle;  it  is  required  to 
describe  the  circle  of  which  it  is  the  segment. 

Bisect  (10. 1.)  AC  in  D,  and  from  the  point  D  draw(l  1.1.)  DB 
at  right  angles  to  AC,  and  join  AB;  first,  let  the  angles  ABD, 
BAD,  be  equal  to  one  another;  then  the  straight  line  BD  is  equal 
(6.  1.)  to  DA,  and  therefore  to  DC,  and  because  the  three  straight 
lines  DA,  DB,  DC,  are  all  equal;  D  is  the  centre  of  the  circle 
(9.  3.):  from  the  centre  D,  at  the  distance  of  any  of  the  three  DA, 
DB,  DC,  describe  a  circle;  this  shall  pass  through  the  other 
points;  and  the  circle  of  which  ABC  is  a  segment  is  described: 
and  because  the  centre  D  is  in  AC;  the  segment  ABC  is  a  semi- 


B 


B 


B 


~JC 


circle:  but  if  the  angles  ABD,  BAD  are  not  equal  to  one  another, 
at  the  point  A,  in  the  straight  line  AB,  make  (23.  1.)  the  angle 
BAE  equal  to  the  angle  ABD,  and  produce  BD,  if  necessary,  to 
E,  and  join  EC:  and  because  the  angle  ABE  is  equal  to  the  angle 
BAE,  the  straight  line  BE  is  equal  (6.  1.)  to  EA;  and  because 
AD  is  equal  to  DC,  and  DE  common  to  the  triangles  ADE,  CDE, 
the  two  sides  AD,  DE  are  equal  to  the  two  CD,  DE,  each  to  each; 
and  the  angle  ADE  is  equal  to  the  angle  CDE,  for  each  of  them 
is  a  right  angle;  therefore  the  base  AE  is  equal  (4.  1.)  to  the  base 
EC:  but  AE  was  shown  to  be  equal  to  EB,  wherefore  also  BE  is 
equal  to  EC:  and  the  three  straight  lines  AE,  EB,  EC  are  there- 
fore equal  to  one  another;  wherefore  (9.  3.)  E  is  the  centre  of  the 
circle.  From  the  centre  E,  at  the  distance  of  any  of  the  three 
AE,  EB,  EC,  describe  a  circle,  this  shall  pass  through  the  other 


See  Note. 


72 


THE  ELEMENTS  OF  EUCLID. 


BOOK  III. 


points;  and  the  circle  of  which  ABC  is  a  segment  is  described: 
and  it  is  evident,  that  if  the  angle  ABD  be  greater  than  the  angle 
BAD,  the  centre  E  falls  without  the  segment  ABC,  which  there- 
fore is  less  than  a  semicircle;  but  if  the  angle  ABD  be  less  than 
BAD,  the  centre  E  falls  within  the  segment  ABC,  which  is  there- 
fore greater  than  a  semicircle:  wherefore  a  segment  of  a  circle 
being  given,  the  circle  is  described  of  which  it  is  a  segment. 
Which  was  to  be  done. 

PROP,  .XXVI.  THEOR. 

In  equal  circles,  equal  angles  stand  upon  equal  cir- 
cumferences, whether  they  be  at  the  centres  or  circum- 
ferences. 

Let  ABC,  DEF  be  equal  circles,  and  the  equal  angles  BGC, 
EHF  at  their  centres,  and  BAC,  EDF  at  their  circumferences: 
the  circumference  BKC  is  equal  to  the  circumference  ELF. 

Join  BC,  EF;  and  because  the  circles  ABC,  DEF  are  equal, 
the  straight  lines  drawn  from  their  centres  are  equal:  therefore 
the  two  sides  BG,  GC  are  equal  to  the  two  EH,  HF;  and  the  an- 

D 


C  E 


gle  at  G  is  equal  to  the  angle  at  H;  therefore  the  base  BC  is 
equal  (4.  1.)  to  the  base  EF;  and  because  the  angle  at  A  is  equal 
to  the  angle  at  D,  the  segment  BAC  is  similar  (l  1  def.  3.)  to  the 
segment  EDF;  and  they  are  upon  equal  straight  lines  BC,  EF; 
but  similar  segments  of  circles  upon  equal  straight  lines  are  equal 
(24.  3.)  to  one  another;  therefore  the  segment  BAC  is  equal  to 
the  segment  EDF;  j)ut  the  whole  circle  ABC  is  equal  to  the  whole 
EDF;  therefore  the  remaining  segment  BKC  is  equal  to  the  re- 
maining segment  ELF,  and  the  circumference  BKC  to  the  cir- 
cumference ELF.     Wherefore,  in  equal  circles,  Sec.  Q.  E.  D. 


PROP.  XXVII.  THEOR. 

In  equal  circles,  the  angles  which  stand  upon  equal 
circumferences  are  equal  to  one  another,  whether  they 
be  at  the  centres  or  circumferences. 

Let  the  angles  BGC,  EHF  at  the  centres,  and  BAC,  EDF  at 
the  circumferences  of  the  equal  circles  ABC,  DEF  stand  upon  the 


t, 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


73 


equal  circumferences  BC,  EFj  the  angle  BGC  is  equal  to  the  an- 
gle EHF,  and  the  angle  BAG  to  the  angle  EDF. 

If  the  angle  BGC  be  equal  to  the  angle  EHF,  it  is  manifest 
(20.  3.)  that  the  angle  BAG  is  also  equal  to  EDF:  but,  if  not,  one 

A 


of  them  is  the  greater;  let  BGG  be  the  greater:  and  at  the  point 
G,  in  the  straight  line  BG,  make  (23.  1.)  the  angle  BGK  equal  to 
the  angle  EHF;  but  equal  angles  stand  upon  equal  circumferences 
(26.  3.)  when  they  are  at  the  centre;  therefore  the  circumference 
BK  is  equal  to  the  circumference  EF:  but  EF  is  equal  to  BG; 
therefore  also  BK  is  equal  to  BC,the  less  to  the  greater,  which  is 
impossible:  therefore  the  angle  BGG  is  not  unequal  to  the  angle 
EHF;  that  is,  it  is  equal  to  it:  and  the  angle  at  A  is  half  of  the 
angle  BGG,  and  the  angle  at  D  half  of  the  angle  EHF:  therefore 
the  angle  at  A  is  equal  to  the  angle  at  D.  Wherefore,  in  equal 
circles,  &c.  Q.  E.  D. 

PROP.  XXVHI.  THEOR. 

In  equal  circles,  equal  straight  lines  cut  off  equal  cir- 
cumferences, the  greater  equal  to  the  greater,  and  the 
less  to  the  less. 

Let  ABG,  DEF  be  equal  circles,  and  BG,  EF  equal  straight 
lines  in  them,  which  cut  off  the  two  greater  circumferences 
BAG,  EDF,  and  the  two  less  BGG,  EHIE';  the  greater  BAG  is 
equal  to  the  greater  EDF,  and  the  less  BGG  to  the  less  EHF. 

Take  (l.  3.)  K,  L,  the  centres  of  the  circles,  and  join  BK,  KG, 
EL,  LF:   and  because  the   circles  are   equal,  the  straight   lines 

A  D 


G  H 

from  their   centres  are  equal:    therefore  BK,  KG  are    equal  to 
EL,  LF;  and  the  base  BG  is  equal  to  the  base  EF;  therefore  the 
angle  BKG  is  equal  (8.  1.)  to  the  angle  ELF:  but  equal  angles 
K 


74  THE  ELEMENTS  OF  EUCLID.  BOOK  III. 

stand  upon  equal  (26.  3.)  circumferences,  when  they  are  at  the 
centres^  therefore  the  circumference  BGC  is  equal  to  the  circum- 
ference EHF.  But  the  whole  circle  ABC  is  equal  to  the  whole 
EDF  ;  the  remaining  part  therefore  of  the  circumference,  viz. 
BAG,  is  equal  to  the  remaining  part  EDF.  Therefore,  in  equal 
circles,  8cc.  Q.  E.  D. 

PROP.  XXIX.  THEOR. 

In  equal  circles  equal  circumferences  are  subtended 
by  equal  straight  lines. 

Let  ABC,  DEF  be  equal  circles,  and  let  the  circumferences 
BGC,  EHF  also  be  equal ;  and  join  BC,  EF:  the  straight  line  BC 
is  equal  to  the  straight  line  EF. 

Take  (1.  3.)  K,  L,  the  centres  of  the  circles,  and  join  BK,  KC, 
EL,  LF:  and  because  the  circumference  BGC  is  equal  to  the  cir- 

A  D 


G  PI 

.-^ 

cumference  EHF,  the  angle  BKC  is  equal  (27.  3.)  to  the  angle 
ELF  :  and  because  the  circles  ABC,  DEF  are  equal,  the  straight 
lines  from  their  centres  are  equal:  therefore  BK,  KC  are  equal 
to  EL,  LF,  and  they  contain  equal  angles:  therefore  the  base  BC 
is  equal  (4.  1.)  to  the  base  EF.  Therefore,  in  equal  circles,  Sec. 
Q.  E.  D. 

PROP.  XXX.  PROB. 

To  bisect  a  given  circumference,  that  is,  to  divide  it 
into  two  equal  parts. 

Let  ADB  be  the  given  circumference,  it  is  required  to  bisect 
it. 

Join  AB,  and  bisect  (10.  1.)  it  in  C ;  from  the  point  C  draw 
CD  at  right  angles  to  AB,  and  join  AD,  DB:  the  circumference 
ADB  is  bisected  in  the  point  D. 

Because  AC  is  equal  to  CB,  and  CB  common  to  the  triangles 
ACD,  BCD,  the  two  sides  AC,  CD  are  equal  D 

to  the  two  BC,  CD ;  and  the  angle  ACD  is 
equal  to  the  angle  BCD,  because  each  of  them 
is  a  right  angle;  therefore  the  base  AD  is 
equal  (4.1.)  to  the  base  BD:  but  equal  straight  A 
lines  cut  off  equal  (28.  3.)  circumferences,  the  greater  equal  to  the 
greater,  and  the  less  to  the  less,  and  AD,  DB  are  each  of  them 
less  than  a  semicircle;  because  DC  passes  through  the  centre 
(Cor.  1.  3.):  wherefore  the  circumference  AD  is  equal  to  the  cir- 


BOOK  III.  THE  ELEMENTS  OF  EUCLID.  75 

cumference  DB  :  therefore  the  given  circumference  is  bisected  in 
D.     Which  was  to  be  done. 

PROP.   XXXI.  THEOR. 

In  a  circle,  the  angle  in  a  semicircle  is  a  right  angle; 
but  the  angle  in  a  segmennt  greater  than  a  semicircle 
is  less  than  a  right  angle ;  and  the  angle  in  a  segment 
less  than  a  semicircle  is  greater  thaji  a  right  angle. 

Let  ABCD  be  a  circle,  of  which  the  diameter  is  BC,  and  centre 
E;  and  draw  CA,  dividing  the  circle  into  the  segments  ABC, 
ADC,  and  join  BA,  AD,  DC;  the  angle  in  the  semicircle  BAC  is 
a  right  angle;  and  the  angle  in  the  segment  ABC,  which  is  greater 
than  a  semicircle,  is  less  than  a  right  angle;  and  the  angle  in  the 
segment  ADC,  which  is  less  than  a  semicircle,  is  greater  than  a 
right  angle. 

Join  AE,  and  produce  BA  to  F;  and  because  BE  is  equal  to 
EA,  the  angle  EAB  is  equal  (5.  1.)  to  EBA;  also,  because  AE  is 
equal  to  EC,  the  angle  EAC  is  equal  to  F 

EC  A;  wherefore  the  whole  angle  BAC  is 
equal  to  the  two  angles  ABC,  ACB;  but 
FAC,  the  exterior  angle  of  the  triangle 
ABC,  is  equal  (32.  1.)  to  the  tw^o  angles 

ABC,  ACB;  therefore  the  angle  BAC  is      /  /  \  "^   D 

equal  to  the  angle  FAC,  and  each  of  them 
is  therefore  a  right  (10.  def.   1.)   angle;  B| 
wherefore  the  angle  BAC  in  a  semicircle 
is  a  right  angle. 

And  because  the  two  angles  ABC, 
BAC,  of  the  triangle  ABC  are  together 
less  (17.  1.)  than  two  right  angles,  and  that  BAC  is  a  right  angle, 
ABC  must  be  less  than  a  right  angle:  and  therefore  the  angle  in 
a  segment  ABC  greater  than  a  semicircle,  is  less  than  a  right 
angle. 

And  because  ABCD  is  a  quadrilateral  figure  in  a  circle,  any 
two  of  its  opposite  angles  are  equal  (22.  3.)  to  two  right  angles; 
therefore  the  angles  ABC,  ADC  are  equal  to  two  right  angles; 
and  ABC  is  less  than  a  right  angle;  wherefore  the  other  ADC  is 
greater  than  a  right  angle. 

Besides,  it  is  manifest,  that  the  circumference  of  the  greater 
segment  ABC  falls  without  the  right  angle  CAB,  but  the  circum- 
ference of  the  less  segment  ADC  falls  within  the  right  angle 
CAF.  "And  this  is  all  that  is  meant,  when  in  the  Greek  text, 
and  the  translations  from  it,  the  angle  of  the  greater  segment  is 
said  to  be  greater,  and  the  angle  of  the  less  segment  is  said  to  be 
less,  than  a  right  angle." 

Cor.  From  this  it  is  manifest,  that  if  one  angle  of  a  triangle  be 
equal  to  the  other  two,  it  is  a  right  angle,  because  the  angle  ad- 
jacent to  it  is  equal  to  the  same  two;  and  when  the  adjacent  an^ 
gles  are  equal,  they  are  right  angles. 


76 


THE  ELEMENTS  OF  EUCLID. 


BOOK  III. 


PROP.  XXXII.  THEOR. 

If  a  straight  line  touch  a  circle,  and  from  the  point 
of  contact  a  straight  line  be  drawn  cutting  the  circle, 
the  angles  made  by  this  line  with  the  line  touching  the 
circle,  shall  be  equal  to  the  angles  which  are  in  the  al- 
ternate segments  of  the  circle. 

Let  the  straight  line  EF  touch  the  circle  ABCD  in  B,  and  from 
the  point  B  let  the  straight  line  BD  be  drawn,  cutting  the  circle: 
The  angles  which  DB  makes  with  the  touching  line  EF  shall  be 
equal  to  the  angles  in  the  alternate  segments  of  the  circle:  that  is, 
the  angle  FBD  is  equal  to  the  angle  which  is  in  the  segment 
DAB,  and  the  angle  DBE  to  the  angle  in  the  segment  BCD. 

From  the  point  B  draw  (11.  1.)  BA  at  right  angles  to  EF,  and 
take  any  point  C  in  the  circumference  BD;  and  join  AD,  DC,  CB; 
and  because  the  straight  line  EF  touches  the  circle  ABCD  in  the 
point  B,  and  BA  is  drawn  at  right 
angles  to  the  touching  line  from 
the  point  of  contact  B,  the  centre 
of  the  circle  is  (19.  3.)  in  BAj 
therefore  the  angle  ADB  in  a  se- 
micircle is  a  right  (31.  3.)  angle, 
and  consequently  the  other  two 
angles  BAD,  ABD  are  equal  (32. 
1.)  to  a  right  angle:  but  ABF  is 
likewise  a  right  angle;  therefore 
the  angle  ABF  is  equal  to  the  an-         E  B  F 

gles  BAD,  ABD:  take  from  these  equals  the  common  angle  ABD; 
therefore  the  remaining  angle  DBF  is  equal  to  the  angle  BAD, 
which  is  in  the  alternate  segment  of  the  circle:  and  because  ABCD 
is  a  quadrilateral  figure  in  a  circle,  the  opposite  angles  BAD, 
BCD  are  equal  (22.  3.)  to  two  right  angles;  therefore  the  angles 
DBF,  DBE,  being  likewise  equaf  (13.  I.)  to  two  right  angles,  are 
equal  to  the  angles  BAD,  BCD;  and  DBF  has  been  proved  equal 
to  BAD:  therefore  the  remaining  angle  DBE  is  equal  to  the  an- 
gle BCD  in  the  alternate  segment  of  the  circle.  Wherefore,  if  a 
straight  line,  Sec.  Q.  E.  D. 

PROP.  XXXIII.  PROB. 

Upon  a  given  straight  line  to  describe  a  segment  of 
a  circle,  containing  an  angle  equal  to  a  given  rectili- 
neal angle. "^ 

Let  AB  be  the  given  straight  line,  and  the  angle  at  C  the  given 
rectilineal  angle;  it  is  required  to  describe  upon  the  given  straight 
line  AB  a  segment  of  a  circle,  containing  an  angle  equal  to  the 
angle  C. 

^  See  Note. 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


77 


First,  Let  the  angle  at  C  be 
a  right  angle,  and  bisect  (10. 
1.)  AB  in  F,  and  from  the  cen- 
tre F,  at  the  distance  FB,  de- 
scribe the  semicircle  AHB; 
therefore  the  angle  AHB  in  a 
semicircle  is  (31.  3.)  equal  to  the  right  angle  at  C. 

But,  if  the  angle  C  be  not  a  right  angle,  at  the  point  A,  in  the 
straight  line  AB,  make  (23.  1.)  the  angle  BAD  equal  to  the  angle 
C,  and  from  the  point  A  draw  ^ 

(11.  1.)  AE  at  right  angles  to 
AD:  bisect  (10.  l.)AB  in  F,  and 
from  F  draw  (11.  1.)  FG  at  right 
angles  to  AB,  and  join  GB:  and 
because  AF  is  equal  to  FB,  and 
FG  common  to  the  triangles 
AFG,  BFG,  the  two  sides  AF, 
FG  are  equal  to  the  two  BF, 
FG;  and  the  angle  AFG  is  equal 
to  the  angle  BFG;  therefore  the 
base  AG  is  equal  (4.  1.)  to  the 
base  GB;  and  the  circle  described  from  the  centre  G,  at  the  dis- 
tance GA,  shall  pass  through  the  point  B;  let  this  be  the  circle 
AHB:  and  because  from  the  point  A  the  extremity  of  the  dia- 
meter AE,  AD  is  drawn  at  right  angles  to  AE,  therefore  AD  (Cor. 
16.  3.)  touches  the  circle;  and  because  AB  drawn  from  the  point 
of  contact  A  cuts  the  circle,  the  " 
angle  DAB  is  equal  to  the  angle 
in  the  alternate  segment  AHB 
(32.  3.):  but  the  angle  DAB  is 
equal  to  the  angle  C,  therefore 
also  the  angle  C  is  equal  to  the 
angle  in  the  segment  AHB: 
wherefore  upon  the  given  straight 

line  AB  the  segment  AHB  of  a  

circle  is  described  which  contains  an  angle  equal  to  the  given 
angle  at  C.     Which  was  to  be  done. 


PROP.  XXXIV.  PROB. 

To  cut  off  a  segment  from  9,  given  circle  which  shall 
contain  an  angle  equal  to  a  given  rectilineal  angle. 

Let  ABC  be  the  given  circle,  and  D  the  given  rectilineal  angle; 
it  is  required  to  cut  off  a  segment  from  the  circle  ABC  that  shall 
contain  an  angle  equal  to  the  given  angle  D. 

Draw  (17.  3.)  the  straight  line  EF  touching  the  circle  ABC  in 


78 


THE  ELEMENTS  OF  EUCLID. 


BOOK  III. 


the  point  B,  and  at  the  point  B, 

in  the  straight  line  BF,  make 

(23.  1.)  the  angle  FBC  equal  to 

the  angle  D;  therefore,  because 

the  straight  line  EF  touches  the 

circle  ABC,  and  BC  is  drawn 

from  the  point  of  contact  B,  the 

angle  FBC  is  equal  (32.  3.)  to  the 

angle  in  the   alternate  segment 

B  AC  of  the  circle:  but  the  angle  E  B  F 

FBC  is  equal  to  the  angle  D;  therefore  the  angle  in  the  segment 

BAC  is  equal  to  the  angle  D:  wherefore  the  segment  BAC  is  cut 

off  from  the  given  circle  ABC  containing  an  angle  equal  to  the 

given  angle  D.     Which  was  to  be  done. 


PROP.  XXXV.  THEOR. 


If  two  straight  lines  within  a  circle  cut  one  another, 
the  rectangle  contained  by  the  segments  of  one  of  them 
is  equal  to  the  rectangle  contained  by  the  segments  of 
the  other.* 

Let  the  two  straight  lines  AC,  BD,  within  the  circle  ABCD, 
cut  one  another  in  the  point  E:  the  rectangle  contained  by  AE, 
EC  is  equal  to  the  rectangle  contained  by  BE, 
ED. 

If  AC,  BD  pass  each  of  them  through  the 
centre,  so  that  E  is  the  centre;  it  is  evident, 
that  AE,  EC,  BE,  ED,  being  all  equal,  the 
rectangle  AE,  EC  is  likewise  equal  to  the 
rectangle  BE,  ED. 

But  let  one  of  them  BD  pass  through  the  centre,  and  cut  the 
other  AC,  which  does  not  pass  through  the  centre,  at  right  angles, 
in  the  point  E;  then,  if  BD  be  bisected  in  F,  F  is  the  centre  of  the 
circle  ABCD;  join  AF:  and  because  BD,  Avhich  passes  through 
the  centre,  cuts  the  straight  line  AC,  which  does  not  pass  through 
the  centre,  at  right  angles  in  E,  AE,  D 

EC  are  equal  (3.  3.)  to  one  another: 
and  because  the  straight  line  BD,is  cut 
into  two  equal  parts  in  the  point  F,  and 
into  two  unequal  in  the  point  E,  the 
rectangle  BE,  ED  together  with  the 
square  of  EF,  is  equal  (5.  2.)  to  the  . 
square  of  FB;  that  is,  to  the  square  of 
FA;  but  the  squares  of  AE,  EF  are 
equal  (47.  1.)  to  the  square  of  FA; 
therefore  the  rectangle  BE,  ED,  to- 
gether with  the  square  of  EF,  is  equal  to  the  squares  of  AE,  EF: 
take  away  the  common  square  of  EF,  and  the  remaining  rectangle 


Sec  Note. 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


79 


BE,  ED  is  equal  to  the  remaining  square  of  AE;  that  is,  to  the 
rectangle  AE,  EC. 

Nex-t,  let  BD,  which  passes  through  the  centre,  cut  the  other 
AC,  which  does  not  pass  through  the  centre,  in  E,  but  not  at  right 
angles:  then,  as  before,  if  BD  be  bisected  in  F,  F  is  the  centre  of 
the  circle.  Join  AF,  and  from  F  draw  (12.  1.)  FG  perpendicular 
to  AC;  therefore  AG  is  equal  (3.  3.)  to  GC;  wherefore  the  rec- 
tangle AE,  EC,  together  with  the  square  of  EG,  is  equal  (5.  2.)  to 
the  square  of  AG:  to  each  of  these  equals  add  the  square  of  GF; 
therefore  the  rectangle  AE,  EC,  together  with  the  squares  of  EG, 
GF,  is  equal  to  the  squares  of  AG,  GF: 
but  the  squares  of  EG,  GF  are  equal 
(47.  1.)  to  the  square  of  EF,  and  the 
squares  of  AG,  GF  are  equal  to  the 
square  of  AF;  therefore  the  rectangle 
AE,  EC,  together  with  the  square  of  . 
EF,  is  equal  to  the  square  of  AF;  that 
^s,  to  the  square  of  F13:  but  the  square 
of  FB  is  equal  (5.  2.)  to  the  rectangle 
BE,  ED,  together  with  the  square  of  EF:  therefore  the  rectangle 
AE,  EC,  together  with  the  square  of  EF,  is  equal  to  the  rectangle 
BE,  ED,  together  with  the  square  of  EF:  take  away  the  common 
square  of  EF,  and  the  remaining  rectangle  AE,  EC  is  therefore 
equal  to  the  remaining  rectangle  BE,  ED. 

Lastly,  Let  neither  of  the  straight  lines  AC,  BD  pass  through 
the  centre:  take  thecentre  F,  and  through 
E,  the  intersection  of  the  straight  lines 
AC,  DB  draw  the  diameter  GEFH:  and 
because  the  rectangle  AE,  EC  is  equal, 
as  has  been  shown,  to  the  rectangle  GE, 
EH:  and,  for  the  same  reason,  the  rect- 
angle BE,  ED  is  equal  to  the  same  rect- 
angle GE,  EH;  therefore  the  rectangle 
AE,  EC  is  equal  to  the  rectangle  BE, 
ED.  Wherefore,  if  two  straight  lines, 
&c.    Q.  E,  D. 


PROP.  XXXVL  THEOR. 

If  from  any  point  without  a  circle  two  straight  hues 
be  drawn,  one  of  which  cuts  the  circle,  and  the  other 
touches  it;  the  rectangle  contained  by  the  whole  line 
which  cuts  the  circle,  and  the  part  of  it  without  the 
circle,  shall  be  equal  to  the  square  of  the  line  Avhich 
touches  it. 

Let  D  be  any  point  without  the  circle  ABC,  and  DCA,  DB  two 
straight  lines  drawn  from  it,  of  which  DCA  cuts  the  circle,  and 
DB  touches  the  same;  the  rectangle  AD,  DC  is  equal  to  the 
square  of  DB. 


80 


THE  ELEMENTS  OF  EUCLID. 


BOOK  III. 


D 


Either  DCA  passes  through  the  centre,  or  it  does  not;  first,  let 
it  pass  through  the  centre  E,  and  join  EB;  therefore  the  angle 
EBD  is  a  right  (18.  3.)  angle:  and  because 
the  straight  line  AC  is  bisected  in  E,  and 
produced  to  the  point  D,  the  rectangle  AD, 
DC,  together  with  the  square  of  EC,  is 
equal  (6.  2.)  to  the  square  of  ED,  and  CE 
is  equal  to  EB:  therefore  the  rectangle 
AD,  DC,  together  with  the  square  of  EB, 
is  equal  to  the  square  of  ED:  but  the 
square  of  ED  is  equal  (47.  1.)  to  the  squares 
of  EB,  BD  because  EBD  is  a  right  angle: 
therefore  the  rectangle  AD,  DC,  together 
with  the  square  of  EB,  is  equal  to  the 
squares  of  EB  BD:  take  away  the  common 
square  of  EB;  therefore  the  remaining  rect- 
angle AD,  DC  is  equal  to  the  square  of  the 
tangent  DB. 

But  if  DCA  does  not  pass  through  the  centre  of  the  circle 
ABC,  take  (l.  3.)  the  centre  E,  and  draw  EF  perpendicular  (12. 
1.)  to  AC,  and  join  EB,  EC,  ED:  and  because  the  straight  line 
EF,  which  passes  through  the  centre,  cuts  the  straight  line  AC, 
which  does  not  pass  through  the  centre,  at  D 

right  angles,  it  shall  likewise  bisect  (3.  3.) 
it;  therefore  AF  is  equal   to  EC:   and  be- 
cause the  straight  line  AC  is  bisected  in  F, 
and  produced  to  D,  the  rectangle  AD,  DC, 
together  with  the  square  of  EC,  is  equal 
(6.  2.)  to  the  square  of  FD:  to  each  of  these 
equals  add  the  square  of  FE;  therefore  the 
rectangle  AD,  DC,  together  with  the  squares 
of  CF,  FE,  is  equal  to  the  squares  of  DF, 
FE:  but  the  square  of  ED  is  equal  (47.  1.) 
to  the  squares  of  DF,  FE,  because  EFD  is  a 
right  angle:  and  the  square  of  EC  is  equal 
to  the   squares  of  CF,  FE;    therefore  the 
rectangle  AD,  DC,  together  with  the  square  of  EC,  is  equal  to 
the  square  of  ED:  and  CE  is  equal  to  EB;  therefore  the  rectan- 
gle AD,  DC,  together  with  the  square  of  EB,  is  equal  to  the 
square  of  ED:  but  the  squares  of  EB,  BD  are  equal  to  the  square 
(47.  1.)  of  ED,  because  EBD  is  a  right  angle;  therefore  the  rect- 
angle AD,  DC,  together  with  the  square  of  EB,  is  equal  to  the 
squares  of  EB,  BD:  take  away  the  common  square  of  EB:  there- 
fore the  remaining  rectangle  AD,  DC  is  equal  to  the  square  oi 
DB.    Wherefore,  if  from  any  point,  &c.   Q.  E.  D.  , 


■^ 


BOOK  III. 


THE  ELEMENTS  OF  EUCLID. 


81 


Cor.  If  from  any  point  without  a  circle, 
there  be  drawn  two  straight  lines  cutting  it, 
as  AB,  AC,  the  rectangles  contained  by  the 
whole  lines  and  the  parts  of  them  without 
the  circle,  are  equal  to  one  another,  viz.  the  "D 
rectangle  BA,  AE  to  the  rectangle  CA,  AF: 
for  each  of  them  is  equal  to  the  square  of 
the  straight  line  AD  which  touches  the  cir- 
cle. 


PROP.  XXXVII.  THEOR. 

If  from  a  point  without  a  circle  there  be  drawn  two 
straight  hnes,  one  of  which  cuts  the  circle,  and  the 
other  meets  it;  if  the  rectangle  contained  by  the  whole 
line,  which  cuts  the  circle,  and  the  part  of  it  without 
the  circle  be  equal  to  the  square  of  the  line  which 
meets  it,  the  line  which  meets  it  shall  touch  the  cir- 
cle.* 

Let  any  point  D  be  taken  without  the  circle  ABC,  and  from 
it  let  two  straight  lines  DCA  and  DB  be  drawn,  of  which  DCA 
cuts  the  circle,  and  DB  meets  it,  if  the  rectangle  AD,  DC  be 
equal  to  the  square  of  DB;  DB  touches  the  circle. 

Draw  (17.  3.)  the  straight  line  DE  touching  the  circle  ABC, 
find  its  centre  F,  and  join  FE,  FB,  FD;  then  FED  is  a  right  (18. 
3.)  angle:  and  because  DE  touches  the  circle  ABC,  and  DCA 
cuts  it,  the  rectangle  AD,  DC  is  equal  (36.  3.)  to  the  square  of 
DE:  but  the  rectangle  AD,  DC  is,  by  hypothesis,  equal  to  the 
square  of  DB:  therefore  the  square  of  DE  is  equal  to  the  square 
of  DB;  and  the  straight  line  DE  equal  to  the  straight  line  DB; 
and  FE  is  equal  to  FB,  wherefore  DE, 
EF  are  equal  to  DB,  BF;  and  the  base 
FD  is  common  to  the  two  triangles 
DEF,  DBF;  therefore  the  angle  DEF 
is  equal  (8.  1.)  to  the  angle  DBF:  but 
DEF  is  a  right  angle,  therefore  also 
DBF  is  a  right  angle:  and  FB,  if  pro- 
duced, is  a  diameter,  and  the  straight 
line  which  is  drawn  at  right  angles  to 
a  diameter,  from  the  extremity  of  it, 
touches  (16.  3.)  the  circle:  therefore 
DB  touches  the  circle  ABC.  Where- 
fore, if  from  a  point,  &c.  Q.  E.  D. 


*  See  Note. 


■■^:: 


THE 


ELEMENTS  OF  EUCL.ID. 


JBOOK  IV. 


DEFINITION'S. 


I. 

A  RECTILINEAL  figurc  is  Said  to  be  inscribed  in  another  rectilineal 
figure,  when  all  the  angles  of  the  inscribed  figure  are  upon  the 
sides  of  the  figure  in  which  it  is  inscribed,  each 
upon  each.* 

II. 

In  like  manner,  a  figure  is  said  to  be  described 
about  another  figure,  when  all  the  sides  of  the 
circumscribed  figure  pass  through  the  angular 
points  of  the  figure  about  which  it  is  described, 
each  through  each. 


III. 

A  rectilineal  figure  is  said  to  be  inscribed 
in  a  circle,  when  all  the  angles  of  the  in- 
scribed figure  are  upon  the  circumference 
of  the  circle. 


IV. 

A  rectilineal  figure  is  said  to  be  described  about  a  circle,  when 
each   side   of    the   circumscribed   figure 
touches  the  circumference  of  the  circle. 

V. 

In  like  manner,  a  circle  is  said  to  be  in- 
scribed in  a  rectilineal  figure,  when  the 
circumference  of  the  circle  touches  each 
side  of  the  figure. 


*  See  Note. 


BOOK  IV.  THE  ELEMENTS  OF  EUCLID.  83 

VI. 

A  circle  is  said  to  be  described  about  a  rectili- 
neal figure, , when  the  circumference  of  the 
circle  passes  through  all  the  angular  points 
of  the  figure  about  which  it  is  described. 

VII. 

A  straight  line  is  said  to  be  placed  in  a  circle,  when  the  extremities 
of  it  are  in  the  circumference  of  the  circle. 

PROP.  I.  PROB. 

In  a  given  circle  to  place  a  straight  line,  equal  to  a 
given  straight  line  not  greater  than  the  diameter  of  the 
circle. 

Let  ABC  be  the  given  circle,  and  D  the  given  straight  line,  not 
greater  than  the  diameter  of  the  circle. 

Draw  BC  the  diameter  of  the  circle  ABC ;  then,  if  BC  be  equal 
to  D,  the  thing  required  is  done;  for  in  the  circle  ABC  a  straight 

line  BC  is  placed  equal  to  D  5  but, 
if  it  be  not,  BC  is  greater  than  D  5 
make  CE  equal  (3.  1.)  to  D,  and  from 
the  centre  C,  at  the  distance  CE, 
describe  the  circle  AEF,  and  join 
CA :  therefore,  because  C  is  the 
centre  of  the  circle  AEF,  CA  is 
equal  to  CE;  but  D  is  equal  to  CE; 
therefore  D  is  equal  to  CA:  where- 
fore, in  the  circle  ABC,  a  straight 
line  is  placed  equal  to  the  given  straight  line  D,  which  is  not 
greater  than  the  diameter  of  the  circle.  Which  was  to  be  done. 

PROP.  II.  PROB. 

In  a  given  circle  to  inscribe  a  triangle  equiangular  to 
a  given  triangle. 

Let  ABC  be  the  given  circle,  and  DEF  the  given  triangle ; 
it  is  required  to  inscribe  in  the  circle  ABC  a  triangle  equiangular 
to  the  triangle  DEF. 

Draw  (17.  3.)  the  straight  line  GAH  touching  the  circle  in  the 
point  A,  and  at  the  point  A,  in  the  straight  line  AH,  make  (23. 
1.)  the  angle  HAC  equal  to  the  angle  DEF;  and  at  the  point  A, 
in  the  straight  line  AG, 
make  the  angle  GAB 
equal  to  the  angle  DFE,  t^ 

and  join  BC  :  therefore 
because  HAG  touches 
the  circle  ABC,  and  AC  / 

is  drawn  from  the  point 
of    contact,     the     angle       I-  F 

HAC  is  equal  (32.  3.)  to 
the  angle  ABC  in  the 
alternate  segment  of  the 


/ 


\ 


84  THE  ELEMENTS  OF  EUCLID.  BOOK  IV. 

circle:  but  HAC  is  equal  to  the  angle  DEF,  therefore  also  the 
angle  ABC  is  equal  to  DEF;  for  the  same  reason,  the  angle  ACB 
is  equal  to  the  angle  DFE;  therefore  the  remaining  angle  BAG  is 
equal  (32.  I.)  to  the  remaining  angle  EDF:  wherefore  the  triangle 
ABC  is  equiangular  to  the  triangle  DEF,  and  it  is  inscribed  in  the 
circle  ABC.     Which  was  to  be  done. 

PROP.  III.  PROB. 

About  a  given  circle  to  describe  a  triangle  equiangu- 
lar to  a  given  triangle. 

Let  ABC  be  the  given  circle,  and  DEF  the  given  triangle;  it  is 
required  to  describe  a  triangle  about  the  circle  ABC  equiangular 
to  the  triangle  DEF. 

Produce  EF  both  ways  to  the  points  G,  H,  and  find  the  centre 
K  of  the  circle  ABC,  and  from  it  draw  any  straight  line  KB;  at 
the  point  K  in  the  straight  line  KB,  make  (23.  1.)  the  angle  BKA 
equal  to  the  angle  DEG,  and  the  angle  BKC  equal  to  the  angle 
DFH;  and  through  the  points  A,  B,  C,  draw  the  straight  lines 
LAM,  MBN,  NCL  touching  (17.  3.)  the  circle  ABC:  therefore 
because  LM,  MN,  NL  touch  the  circle  ABC  in  the  points  A,  B,  C, 
to  which  from  the  centre  are  drawn  KA,  KB,  KC,  the  angles  at 
the  points  A,  B,  C  are  right  (18.  3.)  angles:  and  because  the  four 
angles  of  the  quadrilateral  figure  AMBK  are  equal  to  four  right 
angles,  for  it  can  be  divided  into  two  triangles:  and  that  two  of 
them  KAM,  KBM  are  right  angles,  the  other  two  AKB,  AMB 
are  equal  to  two  right  y 

angles:  but  the  angles 
DEG,  DEF  are  like- 
wise equal  (13.  1.)  to 
two  right  angles ; 
therefore  the  angles 
AKB,  AMB  are  equal 
to  the  angles  DEG, 

DEF  of  which  AKB       /\  /    \G      E  F       H 

is  equal  to  DEG; 
wherefore  the  remain- 
ing   angle   AMB    is 

equal  to  the  remain-  MB  N 

ing  angle  DEF;  in  like  manner,  the  angle  LNM  may  be  demon- 
strated to  be  equal  to  DFE:  and  therefore  the  remaining  angle 
MLN  is  equal  (32.  1.)  to  the  remaining  angle  EDF:  wherefore 
the  triangle  LMN  is  equiangular  to  the  triangle  DEF:  and  it  is 
described  about  the  circle  ABC.     Which  was  to  be  done. 

PROP.  IV.  PROB. 

To  inscribe  a  circle  in  a  given  triangle.^ 

Let  the  given  triangle  be  ABC;  it  is  required  to  inscribe  a 
circle  in  ABC. 

Bisect  (9.  I.)  the  angles  ABC,  BCA  by  the  straight  lines  BD, 

*  See  Note. 


BOOK  IV. 


THE  ELEMENTS  OF  EUCLID. 


85 


r 


CD  meeting  one  another  in  the  point  D,  from  which  draw  (12.  1.) 
DE,DF,DG  perpendiculars  to  AB,  A 

BC,  CA:  and  because  the  angle  EBD 
is  equal  to  the  angle  FED,  for  the 
angle  ABC  is  bisected  by  BD,  and 
that  the  right  angle  BED  is  equal  to 
the  right  angle  BED,  the  two  tri- 
angles EBD,  EBD  have  two  angles 
of  the  one  equal  to  two  angles  of  the 
other,  and  the  side  BD,  which  is 
opposite  to  one  of  the  equal  angles 
in  each  is  common  to  both;  there- 
fore their  other  sides  shall  be  equal 
(26.  1.);  wherefore  DE  is  equal  to  DF:  for  the  same  reason,  DG 
is  equal  to  DE:  therefore  the  three  straight  lines  DE,  DF,  DG  are 
equal  to  one  another,  and  the  circle  described  from  the  centre  D, 
at  the  distance  of  any  of  them,  shall  pass  through  the  extremities 
of  the  other  two,  and  touch  the  straight  lines,  AB,  BC,  CA,  be- 
cause the  angles  at  the  points  E,  F,  G  are  right  angles,  and  the 
straight  line  which  is  drawn  from  the  extremity  of  a  diameter 
at  right  angles  to  it,  touches  (16.  3.)  the  circle:  therefore  the 
straight  lines  AB,  BC,  CA  do  each  of  them  touch  the  circle,  and 
the  circle  EFG  is  inscribed  in  the  triangle  ABC.  Which  was  to 
be  done. 

PROP.  V.  PROB. 

To  describe  a  circle  about  a  given  triangle.* 

Let  the  given  triangle  be  ABC;  it  is  required  to  describe  a  cir- 
cle about  ABC. 

Bisect  (10.  1.)  AB,  AC  in  the  points  D,  E,  and  from  these 
points  draw  DF,  EF  at  right  angles  (11.  1.)  to  AB,  AC;  DF,  EF 
A  A  A 


0 


produced  meet  one  another:  for,  if  they  do  not  meet,  they  are  pa- 
rallel, wherefore  AB,  AC,  which  are  at  right  angles  to  them,  are 
parallel;  which  is  absurd:  let  them  meet  in  F,  and  join  FA;  also, 
if  the  point  F  be  not  in  BC,  join  BE,  CF;  then,  because  AD  is  equal 
to  DB,  and  DF  common,  and  at  right  angles  to  AB,  the  base  AF 
is  equal  (4.  1.)  to  the  base  FB:  in  like  manner,  it  may  be  shown, 
that  CF  is  equal  to  FA;  and  therefore  BE  is  equal  to  EC;  and  FA, 
FB,  EC  are  equal  to  one  another:  wherefore  the  circle  described 
from  the  centre  F,  at  the  distance  of  one  of  them,  shall  pass 


See  Note. 


86  THE  ELEMENTS  OF  EUCLID.  BOOK  IV. 

through  the  extremities  of  the  other  two,  and  be  describedabout 
the  triajigle  ABC.     Which  was  to  be  done. 

Cor.  And  it  is  manifest,  that  when  the  centre  of  the  circle  falls 
within  the  triangle,  each  of  its  angles  is  less  than  a  right  angle, 
each  of  them  being  in  a  segment  greater  than  a  semicircle;  but, 
when  the  centre  is  in  one  of  the  sides  of  the  triangle,  the  angle 
opposite  to  this  side,  being  in  a  semicircle,  is  a  right  angle;  and, 
if  the  centre  falls  without  the  triangle,  the  angle  opposite  to  the 
side  beyond  which  it  is,  being  in  a  segment  less  than  a  semicircle, 
is  greater  than  a  right  angle:  wherefore,  if  the  given  triangle  be 
acute  angled,  the  centre  of  the  circle  falls  within  it;  if  it  be  a  right 
angled  triangle,  the  centre  is  in  the  side  opposite  to  the  right 
angle;  and,  if  it  be  an  obtuse  angled  triangle,  the  centre  falls  with- 
out the  triangle,  beyond  the  side  opposite  to  the  obtuse  angle. 


PROP.  VI.  PROB. 

To  inscribe  a  square  in  a  given  circle. 

Let  ABCD  be  the  given  circle;  it  is  required  to  inscribe  a  square 
in  ABCD. 

Draw  the  diameters  AC,  BD  at  right  angles  to  one  another; 
and  join  AB,  BC,  CD,  DA;  because  BE  is  equal  to  ED,  for  E  is 
the  centre,  and  that  EA  is  common,  and 
at  right  angles  to  BD;  the  base  BA  is 
equal  (4.  1.)  to  the  base  AD;  and  for  the 
same  reason,  BC,  CD  are  each  of  them 

equal  to  BA  or  AD;  therefore  the  quad-   t>  [X        ^  N)   r^ 

rilateral  figure  ABCD  is  equilateral.     It        '^^ 
is  also  rectangular;  for  the  straight  line 
BD,  being  the   diameter   of  the   circle 
ABCD,  BAD  is  a  semicircle;  wherefore 
the  angle  BAD  is  a  right  (31.  3.)  angle;  C 

for  the  same  reason  each  of  the  angles  ABC,  BCD,  CDA  is  a 
right  angle;  therefore  the  quadrilateral  figure  ABCD  is  rectangu- 
lar, and  it  has  been  shown  to  be  equilateral;  therefore  it  is  a 
square;  and  it  is  inscribed  in  the  circle  ABCD.  Which  was  to 
be  done. 

PROP.  VII.  PROB. 

To  describe  a  square  about  a  given  circle. 

Let  ABCD  be  the  given  circle:  it  is  required  to  describe  a 
square  about  it. 

Draw  two  diameters  AC,  BD,  of  the  circle  ABCD,  at  right 
angles  to  one  another,  and  through  the  points  A,  B,  C,  D,  draw 
(17.  3.)  FG,  GH,  HK,  KF  touching  the  circle;  and  because  FG 
touches  the  circle  ABCD,  and  EA  is  drawn  from  the  centre  E  to 
the  point  of  contact  A,  the  angles  at  A  are  right  (18.  3.)  angles; 
for  the  same  reason,  the  angles  at  the  points  B,  C,  D,  are  right 


BOOK  IV. 


THE  ELEMENTS  OF  EUCLID. 


87 


angles^  and  because  the  angle  AEB  is 
a  right  angle,  as  likewise  is  EBG,  Gil 
is  parallel  (28.  I.)  to  AC;  for  the  same 
reason,  AC  is  parallel  to  FK,  and  in 
like  manner  OF,  HK  may  each  of  them 
be  demonstrated  to  be  parallel  to  BED; 
therefore  the  figures  GK,  GC,  AK,  FB, 
BK  are  parallelograms;  and  GF  is 
therefore  equal  (34.  1.)  to  HK,  and  GH 
to  FK;  and  because  AC  is  equal  to  BD,         H  C 

and  that  AC  is  equal  to  each  of  the  two  GH,  FK;  and  BD  to  each 
of  the  two  GF,  HK:  GH,  FK  are  each  of  them  equal  to  GF  or 
HK;  therefore  the  quadrilateral  figure  FGHK  is  equilateral.  It 
is  also  rectangular:  for  GBEA  being  a  parallelogram,  and  AEB 
a  right  angle,  AGB  (34.  1.)  is  likewise  a  right  angle:  in  the  same 
manner,  it  may  be  shown  that  the  angles  at  H,  K,  F  are  right  an- 
gles; therefore  the  quadrilateral  figure  FGHK  is  rectangular,  and 
it  was  demonstrated  to  be  equilateral:  therefore  it  is  a  square; 
and  it  is  described  about  the  circle  ABCD.  Which  was  to  be 
done. 

PROP.  VIII.  PROB. 

To  inscribe  a  circle  in  a  given  square. 

Let  ABCD  be  the  given  square;  it  is  required  to  inscribe  a 
circle  in  ABCD. 

Bisect  (10.  1 .)  each  of  the  sides  AB,  AD,  in  the  points  F,  E,  and 
through  E  draw  (31.  1.)  EH  parallel  to  AB  or  DC,  and  through 
F  draw  FK  parallel  to  AD  or  BC;  therefore  each  of  the  figures 
AK,  KB,  AH,  HD,  AG,  GC,  BG,  GD  is  a  parallelogram,  and 
their  opposite  sides  are  equal  (34.  1.)  and  because  AD  is  equal 
to  AB,  and  that  AE  is  the  half  of  AD,  and  AF  the  half  of  AB, 
AE  is  equal  to  AF;  wherefore  the  sides         A  ED 

opposite  to  these  are  equal,  viz.  EG  to 
GE:  in  the  same  manner,  it  may  be  de- 
monstrated that  GH,  GK  are  each  of 
them  equal  to  EG  or  GE;  therefore  the 
four  straight  lines  GE,  GF,  GH,  GK,  F 
are  equal  to  one  another;  and  the  circle 
described  from  the  centre  G,  at  the  dis- 
tance of  one  of  them,  shall  pass  through 

the  extremities  of  the  other  three,  and         B  H  C 

touch  the  straight  lines  AB,  BC,  CD,  DA;  because  the  angles  at 
the  points  E,  F,  H,  K  are  right  (29.  1.)  angles,  and  that  the 
straight  line  which  is  drawn  from  the  extremity  of  a  diameter,  at 
right  angles  to  it,  touches  the  circle  (l6.  3.)  therefore  each  of  the 
straight  lines  AB,  BC,  CD,  DA  touches  the  circle,  which  there- 
fore is  inscribed  in  the  square  ABCD.     Which  was  to  be  done. 

PROP  IX.    PROB. 

To  describe  a  circle  about  a  given  square. 

Let  ABCD  be  the  given  square;  it  is  required  to  describe  a 
circle  about  it. 


(       ^ 

A 

V 

J 

K 


88 


THE  ELEMENTS  OF  EUCLID. 


BOOK  IV 


Join  AC,  BD  cutting  one  another  in  E;  and  because  DA  is 
equal  to  AB,  and  AC  common  to  the  triangles  DAC,  BAC,  the 

two  sides  DA,  AC  are  equal  to  the  two  BA,     a        pw 

AC;  and  the  base  DC  is  equal  to  the  base  '^         "^ 

BC;  wherefore  the  angle  DAC  is  equal 
(8.  1.)  to  the  angle  BAC,  and  the  angle  DAB 
is  bisected  by  the  straight  line  AC:  in  the 
same  manner,  it  may  be  demonstrated  that 
the  angles  ABC,  BCD,  CDA  are  severally 
bisected  by  the  straight  lines  BD,  AC;  there- 
fore, because  the  angle  DAB  is  equal  to  the 
angle  ABC,  and  that  the  angle  EAB  is  the  half  of  DAB,  and 
EBA  the  half  of  ABC;  the  angle  EAB  is  equal  to  the  angle  EBA; 
Avherefore  the  side  EA  is  equal  (6.1.)  to  the  side  EB:  in  the  same 
manner,  it  may  be  demonstrated  that  the  straight  lines  EC,  ED 
are  each  of  them  equal  to  EA  or  EB:  therefore  the  four  straight 
lines  EA,  EB,  EC,  ED  are  equal  to  one  another;  and  the  circle 
described  from  the  centre  E,  at  the  distance  of  one  of  them,  shall 
pass  through  the  extremities  of  the  other  three,  and  be  described 
about  the  square  ABCD.     Which  was  to  be  done. 


PROP.  X.  PROB. 

To  describe  an  isosceles  triangle,  having  each  of  the 
angles  at  the  base  double  of  the  third  angle. 

Take  any  straight  line  AB,  and  divide  (11.  2.)  it  in  the  point 
C,  so  that  the  rectangle  AB,  BC  be  equal  to  the  square  of  CA; 
and  from  the  centre  A,  at  the  distance  AB,  describe  the  circle 
BDE,  in  which  place  (l.  4.)  the  straight  line  BD  is  equal  to  AC, 
■which  is  not  greater  than  the  diameter  of  the  circle  BDE;  join 
DA,  DC,  and  about  the  triangle  ADC  describe  (5.  4.)  the  circle 
ACi);  the  triangle  ABD  is  such  as  is  required,  that  is,  each  of 
the  angles  ABD,  ADB  is  double  of  the  angle  BAD. 

Because  the  rectangle  AB,  BC  is  equal  to  the  square  of  AC, 
and  that  AC  is  equal  to  BD,  the  rectangle  AB,  BC  is  equal  to  the 
square  of  BD;  and  because  from 
the  point  B  without  the  circle 
ACD  two  straight  lines  BCA,  BD 
are  drawn  to  the  circumference, 
one  of  which  cuts,  and  the  other 
meets  the  circle,  and  that  the  rect- 
angle AB,  BC  contained  by  the 
whole  of  the  cutting  line,  and  the 
part  of  it  without  the  circle  is 
equal  to  the  square  of  BD  which 
meets  it;  the  straight  line  BD 
touches  (37.  3.)  the  circle  ACD; 
and  because  BD  touches  the  cir- 
cle, and  DC  is  drawn  from  the 
point  of  contact  D,  the  angle  BDC 
is  equal  (32.  3.)  to  the  angle  DAC 


E 


B 


D 


BOOK  IV.  THE  ELEMENTS  OF  EUCLID.  89 

in  the  alternate  segment  of  the  circle;  to  each  of  these  add  the 
angle  CD  A:  therefore  the  whole  angle  BDA  is  equal  to  the  two 
angles  CD  A,  DAC;  but  the  exterior  angle  BCD  is  equal  (32.  1.) 
to  the  angles  CDA,  DAC:  therefore  also  BDA  is  equal  to  BCD; 
but  BDA  is  equal  (5.  1.)  to  the  angle  CBD,  because  the  side  AD 
is  equal  to  the  side  AB;  therefore  CBD,  or  DBA  is  equal  to 
BCD;  and  consequently  the  three  angles  BDA,  DBA,  BCD  are 
equal  to  one  another;  and  because  the  angle  DBC  is  equal  to  the 
angle  BCD,  the  side  BD  is  equal  (6.  l.).to  the  side  DC;  but  BD 
was  made  equal  to  CA;  therefore  also  CA  is  equal  to  CD,  and 
the  angle  CDA  equal  (5.  1.)  to  the  angle  DAC;  therefore  the  an- 
gles CDA,  DAC  together,  are  double  of  the  angle  DAC:  but  BCD 
is  equal  to  the  angles  CDA,  DAC;  therefore  also  BCD  is  double 
of  DAC,  and  BCD  is  equal  to  each  of  the  angles  BDA,  DBA; 
each  therefore  of  the  angles  BDA,  DBA,  is  double  of  the  angle 
DAB;  wherefore  an  isosceles  triangle  ABD  is  described,  having 
each  of  the  angles  at  the  base  double  of  the  third  angle.  Which 
was  to  be  done. 

PROP.  XI.   PROB. 

To  inscribe  an  equilateral  and  equiangular  pentagon 
in  a  given  circle. 

Let  ABCDE  be  the  given  circle;  it  is  required  to  inscribe  an 
equilateral  and  equiangular  pentagon  in  the  circle  ABCDE. 

Describe  (10.  4.)  an  isosceles  triangle  FGH,  having  each  of  the 
angles  at   G,  H.,  double  of  the   angle   at  F;    and   in  the  circle 
ABCDE  inscribe  (2.  4.)  the  triangle  ACD  equiangular  to  the  tri- 
angle FGH,  so  that  the  angle  A 
CAD  be  equal  to  the  angle  at 
F,  and  each  of  the  angles  ACD, 
CDA  equal  to  the  angle  at  G 
or  H;   wherefore  each  of  the 
angles  ACD,  CDA  is  double  of 
the  angle  CAD.     Bisect  (9.  1.) 
the  angles  ACD,  CDA  by  the 
straight  lines  CE,  DB:  and  join 
AB,  BC,  DE,   EA.     ABCDE 
is  the  pentagon  required. 

Because  each  of  the  angles 
ACD,  CDA  is  double  of  CAD,  and  are  bisected  by  the  straight 
lines  CE,  DB,  the  five  angles  DAC,  ACE,  ECD,  CDB,  BDA  are 
equal  to  one  another;  but  equal  angles  stand  upon  equal  (26.  3.) 
circumferences;  therefore  the  five  circumferences  AB,  BC,  CD, 
DE,  EA  are  equal  to  one  another:  and  equal  circumferences  are 
subtenij^d  by  equal  (29.  3.)  straight  lines;  therefore  the  five 
straight  lines  AB,  BC,  CD,  DE,  EA  are  equal  to  one  another. 
Wherefore  the  pentagon  ABCDE  is  equilateral.  It  is  also  equi- 
angular; because  the  circumference  AB  is  equal  to  the  circum- 
ference DE:  if  to  each  be  added  BCD,  the  whole  ABCD  is  equal 
to  the  whole  EDCB:  and  the  angle  AED  stands  on  the  circum- 
ference ABCD,  and  the  angle  BAE  on  the  circumference  EDCB; 
M 


90  THE    ELEMENTS  OF  EUCLID.  BOOK  IV. 

.therefore  the  angle  BAE  is  equal  (27.  3.)  to  the  angle  AED:  for 
the  sanie  reason,  each  of  the  angles  ABC,  BCD,  CDE  is  equal  to 
the  angle  BAE,  or  AED:  therefore  the  pentagon  ABCDE  is 
equiangular^  and  it  has  been  shown  that  it  is  equilateral.  Where- 
fore, in  the  given  circle,  an  equilateral  and  equiangular  pentagon 
has  been  inscribed.     Which  was  to  be  done, 

PROP.  XII.  PROB. 

To  describe  an  equilateral  and  equiangular  pentagon 
about  a  given  circle. 

Let  ABCDE  be  the  given  circle;  it  is  required  to  describe  an 
equilateral  and  equiangular  pentagon  about  the  circle  ABCDE. 

Let  the  angles  of  a  pentagon,  inscribed  in  the  circle,  by  the  last 
proposition,  be  in  the  points  A,-B,C,  D,  E,  so  that  the  circumfe- 
rences AB,  BC,  CD,  DE,  EA  are  equal  (11.  ^.);  and  through  the 
points  A,  B,  C,  D,  E  draw  GH,  HK,  KL,  LM,  MG,  touching  (if. 
3.)  the  circle;  take  the  centre  F,  and  join  FB,  FK,  FC,  FL,  FD: 
and  because  the  straight  line  KL  touches  the  circle  ABCDE  in 
the  point  C,  to  which  FC  is  drawn  from  the  centre  F,  FC  is  per- 
pendicular (18.  3.)  to  KL;  therefore  each  of  the  angles  at  C  is  a 
right  angle:  for  the  same  reason,  the  angles  at  the  points  B,  D 
are  right  angles:  and  because  FCK  is  a  right  angle,  the  square  of 
FK  is  equal  (47.  1.)  to  the  squares  of  FC,  CK:  for  the  same  rea- 
son, the  square  of  FK  is  equal  to  the  squares  of  FB,  BK:  there- 
fore the  squares  of  FC,  CK  are  equal  to  the  squares  of  FB,  BK, 
of  which  the  square  of  FC  is  equal  to  the  square  of  FB;  the  re- 
maining square  of  CK  is  therefore  equal  to  the  remaining  square 
of  BK,  and  the  straight  line  CK  equal  to  BK:  and  because  FB  is- 
equal  to  FC,  and  FK  common  to  the  triangles  BFK,  CFK,  the 
two  BF,  FK  are  equal  to  the  two  CF,  FK;  and  the  base  BK  is 
equal  to  the  base  KC;  therefore  the  angle  BFK  is  equal  (8.  1.)  to 
the  angle  KFC,  and  the  angle  BKF  to  FKC;  wherefore  the  angle 
BFC  is  double  of  the  angle  KFC,  and  BKC  double  of  FKC;  for 
the  same  reason,  the  angle  CFD  is  double  of  the  angle  CFL,  and 
CLD  double-  of  CLF:  and  because  the  circumference  BC  is  equal 
to  the  circumference  CD,  the  angle  BFC  is  equal  (27.  3.)  to  the 
angle  CFD;  and  BFC  is  double  of  the 
angle  KFC,  and  CFD  double  of  CFL; 
therefore  the  angle  KFC  is  equal  to 
the  angle  CFL;  and  the  right  angle 
FCK  is  equal  to  the  right  angle  FCL: 
therefore,  in  the  two  triangles  FKC, 
FLC,  there  are  two  angles  of  one 
equal  to  two  angles  of  the  other,  each 
to  each,  and  the  side  FC,  v/hich  is 
adjacent  to  the  equal  angles  in  each, 
is  common  to  both;  therefore  the 
other  sides  shall  be  equal  (26.  1.)  to  K  C         L 

the  other  sides,  and  the  third  angle  to  the  third  angle:  therefore 
the  straight  line  KC  is  equal  to  CL,  and  the  anr^ie  FKC  to  the 


BOOK  IV. 


THE  ELEMENTS  OF  EUCLID. 


91 


angle  FLC:  and  because  KG  is  equal  to  CL,  KL  is  double  of  KC: 
in  the  same  manner,  it  may  be  shown  that  HK  is  double  of  BK: 
and  because  BK  is  equal  to  KC,  as  was  demonstrated,  and  that 
KL  is  double  of  KC,  and  HK  double  of  BK,  HK  shall  be  equal  to 
KL:  in  like  manner,  it  may  be  shown  that  GH,  GM,  ML  are  each 
of  them  equal  to  HK  or  KL:  therefore  the  pentagon  GHKLM  is 
equilateral.  It  is  also  equiangular^  for,  since  the  angle  FKC  is 
equal  to  the  angle  FLC,  and  that  the  angle  HKL  is  double  of  the 
angle  FKC,  and  KLM  double  of  FLC,  as  was  before  demon- 
strated, the  angle  HKL  is  equal  to  KLM:  and  in  like  manner  it 
may  be  shown,  that  each  of  the  angles  KHG,  HGM,  GML  is 
equal  to  the  angle  HKL  or  KLM:  therefore  the  five  angles  GHK, 
HKL,  KLM,  LMG,  MGH  being  equal  to  one  another,  the  pen- 
tagon GHKLM  is  equiangular:  and  it  is  equilateral,  as  was 
demonstrated  J  .  and  it  is  described  about  the  circle  ABCDE. 
Which  was  to  be  done. 


PROP.  XHL  PROB. 

To  inscribe  a  circle  in  a  given  equilateral  and  equi- 
angular pentagon. 

Let  ABCDE  be  the  given  equilateral  and  equiangular  penta- 
gon: it  is  required  to  inscribe  a  circle  in  the  pentagon  ABCDE. 

Bisect  (9.  1.)  the  angles  BCD,  CDE  by  the  straight  lines  CF, 
DF,  and  from  the  point  F,  in  which  they  meet,  drav/  the  straight 
lines  FB,  F^\.  FE;  thei;^ore,  since  BC  is  equal  to  CD,  and  CF 
common  to  'uie  triangles  BCF,  DCF,  the  two  sides  BC,  CF  are 
equal  to  the  two  DC,  CF;  and  the  angle  BCF  is  equal  to  the 
angle  DCFj  therefore  the  base  BF  is  equal  (4.  1.)  to  the  base  FD, 
and  the  other  angles  to  the  other  angles,  to  which  the  equal  sides 
are  opposite;  therefore  the  angle  CBF  is  equal  to  the  angle  CDF: 
and  because  the  angle  CDE  is  double  of  CDF,  and  that  CDE  is 
equal  to  CBA,  and   CDF   to   CBF;  A 

CBA  is  also  double  of  the  angle  CBF; 
therefore  the  angle  ABF  is  equal  to 
the  angle  CBF;  wherefore  the  angle 
ABC  is  bisected  by  the  straight  line 
BF:  in  the  same  manner  it  may  be 
demonstrated,  that  the  angles  BAE, 
AED  are  bisected  by  the  straight 
lines  AF,  FE:  from  the  point  F  draw 
(12.  1.)  FG,  FH,  FK,  FL,  FM  per- 
pendiculars to  the  straight  lines  AB, 
BC,  CD,  DE,  EA;  and  because  the 
angle  HGF  is  equal  to  KCF,  and  the  C  K     ^    D 

right  angle  FHC  equal  to  the  right  angle  FKC;  in  the  triangles 
FHC,  FKC  there  are  two  angles  of  one  equal  to  two  angles  of  the 
other,  and  the  side  FC,  which  is  opposite  to  one  of  the  equal 
angles  in  each,  is  common  to  both;  therefore  the  other  sides  shall 
be  equal  (26.  1.)  each  to  each;  wherefore  the  perpendicular  FH  is 
equal  to  the  perpendicular  FK:  in  the  same  manner  it  may  be  de- 


92  .      THE  ELEMENTS  OF  EUCLID.  BOOK  IV. 

monstrated  that  FL,  FM,  FG  are  each  of  them  equal  to  FH,  or 
FK;  therefore  the  five  straic^ht  lines  FG,  FH,  FK,  FL,  FM  are 
equal  to  one  another:  wherefore  the  circle  described  frora  the 
centre  F,  at  the  distance  of  one  of  these  five,  shall  pass  through 
the  extremities  of  the  other  four,  and  touch  the  straight  lines  AB, 
BC,  CD,  DE,  EA,  because  the  angles  at  the  points  G,  H,  K,  L,  M 
are  right  angles;  and  that  a  straight  line  drawn  from  the  extre- 
mity of  the  diameter  of  a  circle  at  right  angles  to  it,  touches 
(16.  3.)  the  circle:  therefore  each  of  the  straight  lines  AB,  BC, 
CD,  DE,  EA  touches  the  circle;  wherefore  it  is  inscribed  in  the 
pentagon  ABCDE.    Which  was  to  be  done. 


PROP.  XIV.   PROB. 

To  describe  a  circle  about  a  given  equilateral  and 
equiangular  pentagon. 

Let  ABCDE  be  the  given  equilateral  and  equiangular  penta- 
gon; it  is  required  to  describe  a  circle  about  it. 

Bisect  (9.  1.)  the  angles  BCD,  CDE  by  the  straight  lines  CF, 
FD,  and  from  the  point  F,  in  which  they  meet,  draw  the  straight 
lines  FB,  FA,  FE,  to  the  points  B,  A,  A 

E.  It  may  be  demonstrated,  in  the 
sanne  manner  as  in  the  preceding  propo- 
sition, that  the  angles  CBA,  BAE,  AED 
are  bisected  by  the  straight  lines  FB, 
FA,  FE:  and  because  the  angle  BCD  is 
equal  to  the  angle  CDE,  and  that  FCD 
is  the  half  of  the  angle  BCD,  and  CDF 
the  half  of  CDE;  the  angle  FCD  is 
equal  to  FDC;  wherefore  the  side  CF  is 
equal  (6.  1.)  to  the  side  FD:  in  like  man-  C"^  '       "^D 

ner  it  may  be  demonstrated  that  FB,  FA,  FE  are  each  of  them 
equal  to  FC  or  FD:  therefore  the  five  straight  lines  FA,  FB,  FC, 
FD,  FE  are  equal  to  one  another;  and  the  circle  described  from 
the  centre  F,  at  the  distance  of  one  of  them,  shall  pass  through 
the  extremities  of  the  other  four,  and  be  described  about  the  equi- 
lateral and  equiangular  pentagon  ABCDE.  Which  was  to  be 
done. 

PROP.  XV.  PROB. 

To  inscribe  an  equilateral  and  equiangular  hexagon 
in  a  given  circle."^ 

Let  ABCDEF  be  the  given  circle;  it  is  required  to  inscribe  an 
equilateral  and  equiangular  hexagon  in  it. 

Find  the  centre  G  of  the  circle  ABCDEF,  and  draw  the  dia- 
meter AGD;  and  from  D  as  a  centre,  at  the  distance  DG,  de- 
scribe the  circle  EGCH,  join  EG,  CG,  and  produce  them  to  the 

'  See  Note. 


BOOK  IV. 


THE  ELEMENTS  OF  EUCLID. 


points  B,  F;  and  join  AB,  BC,  CD,  DE,  EF,  FA:  the  hexagon 
ABCDEF,  is  equilateral  and  equiangular. 

Because  JG  is  the  centre  of  the  circle  ABCDEF,  GE  is  equal  to 
GD:  and  because  D  is  the  centre  of  the  circle  EGCH,  DE  is 
equal  to  DG;  wherefore  GE  is  equal  to  ED,  and  the  triangle 
EGD  is  equilateral;  and  therefore  its  three  angles  EGD,  GDE, 
DEG  are  equal  to  one  another,  because  the  angles  at  the  base  of 
an  isosceles  triangle  are  equal  (5.  1.);  and  the  three  angles  of  a 
triangle  are  equal  (32.  1.)  to  two  right  angles;  therefore  the  angle 
EGD  is  the  third  part  of  two  right  angles:  in  the  same  manner 
it  may  be  demonstrated,  that  the   angle  A 

DGC  is  also  the  third  part  of  two  right 
angles:  and  because  the  straight  line  GC 
makes  with  EB  the  adjacent  angles  EGC, 
CGB  equal  (13.  1.)  to  two  right  angles; 
the  remaining  angle  CGB  is  the  third 
part  of  two  right  angles;  therefore  the 
angles  EDG,  DGC,  CGB  are  equal  to  one 
another:  and  to  these  are  equal  (15.  1.) 
the  vertical  opposite  angles  BGA,  AGF, 
FGE:  therefore  the  six  angles  EGD, 
DGC,  CGB,  BGA,  AGF,  FGE  are  equal 
to  one  another:  but  equal  angles  stand 
upon  equal  (26.  3.)  circumferences;  there- 
fore the  six  circumferences  AB,  BC,  CD, 
DE,  EF,  FA  are  equal  to  one  another:  and  equal  circumferences 
are  subtended  by  equal  (29.  3.)  straight  lines;  therefore  the  six 
straight  lines  are  equal  to  one  another,  and  the  hexagon  ABCDEF, 
is  equilateral.  It  is  also  equiangular;  for,  since  the  circumfer- 
ence AE  is  equal  to  ED,  to  each  of  these  add  the  circumference 
ABCD:  therefore  the  whole  circumference  FABCD  shall  be 
equal  to  the  whole  EDCBA:  and  the  angle  FED  stands  upon  the 
circumference  FABCD," and  the  angle  AFE  upon  EDCBA;  there- 
fore the  angle  AFE  is  equal  to  FED:  in  the  same  manner  it  may 
be  demonstrated  that  the  other  angles  of  the  hexagon  ABCDEF 
are  each  of  them  equal  to  the  angle  AFE  or  FED;  therefore  the 
hexagon  is  equiangular;  and  it  is  equilateral,  as  was  shown;  and 
it  is  inscribed  in  the  given  circle  ABCDEF.  Which  was  to  be 
done. 

Cor.  From  this  it  is  manifest,  that  the  side  of  the  hexagon  is 
equal  to  the  straight  line  from  the  centre,  that  is,  to  the  semidia- 
meter  of  the  circle. 

And  if  through  the  points  A,  B,  C,  D,  E,  F  there  be  draAvn 
straight  lines  touching  the  circle,  an  equilateral  and  equiangular 
hexagon  shall  be  described  about  it,  which  may  be  demonstrated 
from  what  has  been  said  of  the  pentagon;  and  likewise  a  circle 
may  be  inscribed  in  a  given  equilateral  and  equiangular  hexagon, 
and  circumscribed  about  it,  by  a  method  like  to  that  used  for  the 
pentagon. 


94 


THE  ELEMENTS  OF  EUCLID. 


ROOK  IV 


PROP.  XVI.  PROB. 

To  inscribe  an  equilateral  and  equiangular  quindeca- 
gon  in  a  given  circle.* 

Let  ABCD  be  the  given  circle;  it  is  required  to  inscribe  an 
equilateral  and  equiangular  quindecagon  in  the  circle  ABCD. 

Let  AC  be  the  side  of  an  equilateral  triangle  inscribed  (2.  4.) 
in  the  circle,  and  AB  the  side  of  an  equilateral  and  equiangular 
pentagon  inscribed  (H.  4.)  in  the  same:  therefore,  if  such  equal 
p^arts  as  the  whole  circumference  ABCDF  contains  fifteen,  the 
circumference  ABC,  being  the  third  A 

part  of  the  whole,  contains  five;  and 
the  circumference  AB,  which  is  the 
fifth  part  of  the  whole,  contains  three; 
therefore  BC  their  difference  contains 
two  of  the  same  parts:  bisect  (H.  4.) 
BC  in  E;  therefore  BE,  EC  are,  each 
of  them,  the  fifteenth  part  of  the  whole 
circumference  ABCD:  therefore,  if 
the  straight  lines  BE,  EC  be  drawn, 
and  straight  lines  equal  to  them  be 
placed  (1.4.)  around  in  the  whole  circle,  an  equilateral  and  equi- 
angular quindecagon  shall  be  inscribed  in  it.  Which  was  to  be 
done. 

And  in  the  same  manner  as  was  done  in  the  pentagon,  if  through 
the  points  of  division  made  by  inscribing  the  quindecagon,  straight 
lines  be  drawn  touching  the  circle,  an  equilateral  and  equiangular 
quindecagon  shall  be  described  about  it:  and  likewise  as  in  the 
pentagon,  a  circle  may  be  inscribed  in  a  given  equilateral  and 
equiangular  quindecagon,  and  circumscribed  about  it. 


*  See  Note. 


THE 


ELEMENTS  OF  EUCLID. 


BOOK  V. 


DEFiiriTioxrs. 

I. 

A  LESS  magnitude  is  said  to  be  a  part  of  a  greater  magnitude, 
when  the  less  measures  the  greater,  that  is,  *  when  the  less  is 
contained  a  certain  number  of  times  exactly  in  the  greater.' 

II. 

A  greater  magnitude  is  said  to  be  a  multiple  of  a  less,  when  the 
greater  is  measured  by  the  less,  that  is,  *  when  the  greater  con- 
tains the  less  a  certain  number  of  times  exactly.' 

III. 

'  Ratio  is  a  mutual  relation  of  two  magnitudes  of  the  same  kind 
to  one  another,  in  respect  of  quantity.'* 

IV. 

Magnitudes  are  said  to  have  a  ratio  to  one  another,  when  the  less 
can  be  multiplied  so  as  to  exceed  the  other. 

V. 

The  first  of  four  magnitudes  is  said  to  have  the  same  ratio  to  the 
second,  which  the  third  has  to  the  fourth,  when  any  equimulti- 
ples whatsoever  of  the  first  and  third  being  taken,  and  any  equi- 
multiples whatsoever  of  the  second  and  fourth^  if  the  multiple 
of  the  first  be  less  than  that  of  the  second,  the  multiple  of  the 
third  is  also  less  than  that  of  the  fourthj  or,  if  the  multiple  of 
the  first  be  equal  to  that  of  the  second,  the  multiple  of  the  third 
is  also  equal  to  that  of  the  fourth;  or,  if  the  multiple  of  the  first 
be  greater  than  that  of  the  second,  the  multiple  of  the  third  is 
also  greater  than  that  of  the  fourth. 

VI. 

Magnitudes. which  have  the  same  ratio  are  called  proportionals. 

"^  See  Note. 


96    '  THE  ELEMEMTS  OF  EUCLID.  BOOK  V. 

N.  B.  '  When  four  magnitudes  are  proportionals,  it  is  usually 
expressed  by  saying,  the  first  is  to  the  second,  as  the  third  to 
the  fourth.' 

VII. 
When  of  the  equimultiples  of  four  magnitudes  (taken  as  in  the 
fifth  definition)  the  multiple  of  the  first  is  greater  than  that  of 
the  second,  but  the  multiple  of  the  third  is  not  greater  than  the 
multiple  of  the  fourth;  then  the  first  is  said  to  have  to  the 
second  a  greater  ratio  than  the  third  magnitude  has  to  the 
fourth;  and,  on  the  contrary,  the  third  is  said  to  have  to  the 
fourth  a  less  ratio  than  the  first  has  to  the  second. 

VIII. 
'  Analogy,  or  proportion,  is  the  similitude  of  ratios.' 

IX. 
Proportion  consists  in  three  terms  at  least. 

X. 

When  three  magnitudes  are  proportionals,  the  first  is  said  to 
have  to  the  third  the  duplicate  ratio  of  that  which  it  has  to  the 
second. 

XI. 

When  four  magnitudes  are  continual  proportionals,  the  first  is 
said  to  have  to  the  fourth  the  triplicate  ratio  of  that  which  it 
has  to  the  second,  and  so  on,  quadruplicate,  8cc.  increasing  the 
denomination  still  by  unity,  in  any  number  of  proportionals. 

Definition  A,  to  wit,  of  compound  ratio. 

When  there  are  any  number  of  magnitudes  of  the  same  kind,  the 
first  is  said  to  have  to  the  last  of  them  the  ratio  compounded 
of  the  ratio  which  the  first  has  to  the  second,  and  of  the  ratio 
which  the  second  has  to  the  third,  and  of  the  ratio  which  the 
third  has  to  the  fourth,  and  so  on  unto  the  last  magnitude. 

For  example,  if  A,  B,  C,  D  be  four  magnitudes  of  the  same  kind, 
the  first  A  is  said  to  have  to  the  last  D  the  ratio  compounded 
of  the  ratio  of  A  to  B,  and  of  the  ratio  of  B  to  C,  and  of  the 
ratio  of  C  to  D;  or,  the  ratio  of  A  to  D  is  said  to  be  com- 
pounded of  the  ratios  of  A  to  B,  B  to  C,  and  C  to  D. 

And  if  A  have  to  B  the  same  ratio  which  E  has  to  F;  and  B  to 
C,  the  same  ratio  that  G  has  to  H;  and  C  to  D,  the  same  that 
K  has  to  L;  then,  by  this  definition,  A  is  said  to  have  to  D  the 
ratio  compounded  of  ratios  which  are  the  same  with  the  ratios 
of  E  to  F,  G  to  H,  and  K  to  L:  and  the  same  thing  is  to  be  un- 
derstood when  it  is  more  briefly  expressed,  by  saying  A  has  to 
D  the  ratio  compounded  of  the  ratios  of  E  to  F,  G  to  H,  and 
K  to  L. 

In  like  manner  the  same  things  being  supposed,  if  M  have  to  N 
the  same  ratio  which  A  has  to  D:  then  for  shortness'  sake,  M 
is  said  to  have  to  N,  the  ratio  compounded  of  the  ratios  of  E  to 
F,  G  to  H,  and  K  to  L. 

XII. 

In  proportionals,  the  antecedent  terms  are  called  homologous  to 
one  another,  as  also  the  consequents  to  one  another. 


BOOK  V.  THE  ELEMENTS  OF  EUCLID.  97 

*  Geometers  make  use  of  the  following-  technical  words  to  signify 
certain  ways  of  changing  either  the  order  or  magnitude  of  pro- 
portionals, so  as  that  they  continue  still  to  be  proportionals.' 

XIII. 

Permutando,  or  alternando,  by  permutation,  or  alternately;  this 
word  is  used  when  there  are  four  proportionals,  and  it  is  in- 
ferred, that  the  first  has  the  same  ratio  to  the  third,  which  the 
second  has  to  the  fourth;  or  that  the  first  is  to  the  third,  as  the 
second  to  the  fourth;  as  is  shown  in  the  16th  prop,  of  this  5th 
book.* 

XIV. 

Invertendo,  by  inversion;  when  there  are  four  proportionals,  and 
it  is  inferred,  that  the  second  is  to  the  first,  as  the  fourth  to  the 
third.     Prop.  B.  book  5. 

XV. 

Componendo,  by  composition;  when  there  are  four  proportionals, 
and  it  is  inferred,  that  the  first,  together  with  the  second,  is  to 
the  second,  as  the  third,  together  with  the  fourth,  is  to  the 
fourth.     18th  Prop,  book  5. 

XVI. 

Dividendo,  by  division;  when  there  are  four  proportionals,  and  it 
is  inferred,  that  the  excess  of  the  first  above  the  second,  is  to 
the  second,  as  the  excess  of  the  third  above  the  fourth,  is  to 
the  fourth.     17th  Prop,  book  5. 

XVII. 

Convertendo,  by  conversion;  when  there  are  four  proportionals, 
and  it  is  inferred,  that  the  first  is  to  its  excess  above  the  se- 
cond, as  the  third  to  its  excess  above  the  fourth.  Prop.  E. 
book  5. 

XVIII. 

Ex  aequali  (sc.  distantia),  or  ex  aequo,  from  equality  of  distance 
when  there  is  any  number  of  magnitudes  more  than  two,  and 
as  many  others,  so  that  they  are  proportionals  when  taken  two 
and  two  of  each  rank,  and  it  is  inferred,  that  the  first  is  to  the 
last  of  the  first  rank  of  magnitudes,  as  the  first  is  to  the  last  of 
the  others:  'Of  this  there  are  the  two  following  kinds,  which 
arise  from  the  different  order  in  which  the  magnitudes  are 
taken  two  and  two.' 

XIX. 

Ex  aequali,  from  equality;  this  term  is  used  simply  by  itself,  when 
the  first  magnitude  is  to  the  second  of  the  first  rank,  as  the  first 

*  S«e  Note. 

N 


98  THE  ELEMENTS  OF  EUCLID.  BOOK  V. 

to  the  second  of  the  other  rank:  and  as  the  second  is  to  the 
third  of  the  first  rank,  so  is  the  second  to  the  third  of  the  other: 
and  so  on  in  order,  and  the  inference  is  as  mentioned  in  the 
preceding  definition;  whence  this  is  called  ordinate  proportion. 
It  is  demonstrated  in  22d  Prop,  book  5. 

XX. 

Ex  aequali,  in  proportione  perturbata,  sen  inordinata;  from  equa- 
lity, in  perturbata  or  disorderly  proportion;*  this  term  is  used 
when  the  first  magnitude  is  to  the  second  of  the  first  rank,  as 
the  last  but  one  is  to  the  last  of  the  second  rank;  and  as  the 
second  is  to  the  third  of  the  first  rank,  so  is  the  last  but  two  to 
the  last  but  one  of  the  second  rank;  and  as  the  third  is  to  the 
fourth  of  the  first  rank,  so  is  the  third  from  the  last  to  the  last 
but  two  of  the  second  rank:  and  so  on  in  a  cross  order:  and  the 
inference  is  as  in  the  18th  definition.  It  is  demonstrated  in  the 
23d  Prop,  of  book  5. 


AXIOMS. 

I. 

Equimultiples  of  the  same,  or  of  equal  magnitudes,  are  equal  to 
one  another. 

II. 

Those  magnitudes  of  which  the  same,  or  equal  magnitudes  are 
equimultiples,  are  equal  to  one  another. 

III. 

A  multiple  of  a  greater  magnitude  is  greater  than  the  same  mul- 
tiple of  a  less. 

IV. 

That  magnitude  of  which  a  multiple  is  greater  than  the  same 
multiple  of  another,  is  greater  than  that  other  magnitude. 

PROP.  I.  THEOR. 

If  any  number  of  magnitudes  be  equimultiples  of  as 
many,  each  of  each;  what  multiple  soever  any  one  of 
them  is  of  its  part,  the  same  multiple  shall  all  the  first 
magnitudes  be  of  all  the  other. 

Let  any  number  of  magnitudes  AB,  CD  be  equimultiples  of  as 
many  others  E,  F,  each  of  each;  whatsoever  multiple  AB  is  of 
E,  the  same  multiple  shall  AB  and  CD  together  be  of  E  and  F 
together. 

Because  AB  is  the  same  multiple  of  E  that  CD  is  of  F,  as  many 
magnitudes  as  are  in  AB  equal  to  E,  so  many  are  there  in  CD 

*  4  Prop.  lib.  2.     Archimedis  de  sphssra  et  cylindro. 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


99 


equal  to  F.  Divide  AB  into  magnitudes  equal 
to  E,  viz.  AG,  GB,-  and  CD  into  CH,  HD  equal 
each  of  them  to  F:  the  number  therefore  of  the 
magnitudes  CH,  HD  shall  be  equal  to  the  num-      ^  E 

ber  of  the  others,  AG,  GB;  and  because  AG  is 
equal  to  E,  and  CH  to  F,  therefore  AG  and  CH 
together  are  equal  to  (Ax  2.  5.)  E  and  F  toge-      B 
ther:  for  the  same  reason,  because  GB  is  equal 
to  E,  and  HD  to  F,-  GB  and  HD  together  are      C 
equal  to  E   and    F    together.     Wherefore,  as 
many  magnitudes  as  are  in  AB  equal  to  E,  so     H" 
many  are  there  in  AB,  CD  together  equal  to  E 
and  F  together.     Therefore,  whatsoever  multi- 
ple AB  is  of  E,  the  same  multiple  is  AB  and      D 
CD  together  of  E  and  F  together. 

Therefore,  if  any  magnitudes,  how  many  soever,  be  equimulti- 
ples of  as  many,  each  of  each,  whatsoever  multiple  any  one  of 
them  is  of  its  part,  the  same  multiple  shall  all  the  first  magni- 
tudes be  of  all  the  other:  '  For  the  same  demonstration  holds  in 
'  any  number  of  magnitudes,  which  was  here  applied  to  two.' 
Q.  E.  D. 

PROP.  H.  THEOR. 

If  the  first  magnitude  be  the  same  multiple  of  the 
second  that  the  third  is  of  the  fourth,  and  the  fifth  the 
same  multiple  of  the  second  that  the  sixth  is  of  the 
fourth ;  then  shall  the  first  together  with  the  fifth  be 
the  same  multiple  of  the  second,  that  the  third  together 
with  the  sixth  is  of  the  fourth. 

Let  AB  the  first,  be  the  same  multiple  of  C  the  second,  that 
DE  the  third  is  of  F  the  fourth;  and  BG  the  fifth,  the  same  mul- 
tiple of  C  the  second,  that  EH  the 
sixth  is  of  F  the  fourth:  then  is  AG         A  D 

the  first,  together  with  the  fifth,  the 
same  multiple  of  C  the  second,  that 
DH  the  third,  together  with  the 
sixth,  is  of  F  the  fourth.  ^  E 

Because  AB  is  the  same  multiple 
of  C,  that  DE  is  of  F;  there  are  as 
many  magnitudes  in  AB  equal  to 
C,  as  there  are  in  DE  equal  to  F:         q  H 

in  like  manner,  as  many  as   there 

are  in  BG  equal  to  C,  so  many  are  there  in  EH  equal  to  F:  as 
many,  then,  as  are  in  the  whole  AG  equal  to  C,  so  many  are  there 
in  the  whole  DH  equal  to  F:  therefore  AG  is  the  same  multiple 


F 


100 


THE  ELEMENTS  OF  EUCLID. 


BOOK  V 


of  C,  that  DH  is  of  F;  that  is,  AG  A 

the  first   and  fifth  together,   is  the 

same  multiple  of  the  second  C,  that 

DH  the  third  and  sixth  together,  is 

of  the  fourth  F.     H",  therefore,  the      B- 

first  be  the   same  multiple,  Sec.  Q. 

E.  D. 

CoR.    *  From  this  it  is  plain,  that, 

*  if  any  number  of  magnitudes  AB,      G. 
'  BG,  GH,  be  multiples  of  another  C; 

*  and  as  many  DE,  EK,  KL,  be  the 
'  same  multiples  of  F,  each  of  each; 

*  the  whole  of  the  first,  viz.  AH,  is 
'the  same  multiple  of  C,  that  the 

*  whole  of  the  last,  viz.  DL,  is  of  F.' 


D 

E- 


K- 


H      C 


H 


PROP.  HI.  THEOR. 

If  the  first  be  the  same  multiple  of  the  second,  which 
the  third  is  of  the  fourth;  and  if  of  the  first  and  third 
there  be  taken  equimultiples,  these  shall  be  equimulti- 
ples, the  one  of  the  second,  and  the  other  of  the  fourth. 

Let  A  the  first,  be  the  same  multiple  of  B  the  second,  that  C 
the  third  is  of  D  the  fourth;  and  of  A,  C  let  the  equimultiples 
EF,  GH  be  taken;  then  EF  is  the  same  multiple  of  B  that  GH 
is  of  D. 

Because  EF  is  the  same  multiple  of  A  that  GH  is  of  C,  there 
are  as  many  magnitudes  in  EF  equal  to  A;  as  are  in  GH  equal 
to  C:  let  EF  be  divided  into  the 
magnitudes  EK,  KF,  each  equal 
to  A,  and  GH  into  GL,  LH,  each 
equal  to  C :  the  number  therefore 
of  the  magnitudes  EK,  KF,  shall 
be  equal  to  the  number  of  the 
others  GL,  LH:  and  because  A  is 
the  same  multiple  of  B,  that  C  is 
of  D,  and  that  EK  is  equal  to  A, 
and  GL  to  C;  therefore  EK  is  the 
same  multiple  of  B,  that  GL  is  of 
D:  for  the  same  reason,  KF  is  the 
same  multiple  of  B,  that  LH  is  of 
D;  and  so,  if  there  be  more  parts 
in  EF,  GH  equal  to  A,  C :  because, 

therefore,  the  first  EK  is  the  same  multiple  of  the  second  B,  which 
the  third  GL  is  of  the  fourth  D,  and  that  the  fifth  KF  is  the  same 
multiple  of  the  second  B,  which  the  sixth  LH  is  of  the  fourth  D; 
EF  the  first  together  with  the  fifth,  is  the  same  multiple  (2.  5.) 
of  the  second  B,  which  GH  the  third,  together  with  the  sixth,  is 
of  the  fourth  D.     If,  therefore,  the  first,  See.  Q.  E,  D. 


K— 


E 


B     G     C      D 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


101 


PROP.  IV.  THEOR. 

If  the  first  of  four  magnitudes  has  the  same  ratio  to 
the  second  which  the  third  has  to  the  fourth,  then  any 
equimultiples  whatever  of  the  first  and  third  shall  have 
the  same  ratio  to  any  equimultiples  of  the  second  and 
fourth,  viz.  'the  equimultiple  of  the  first  shall  have  the 
'  same  ratio  to  that  of  the  second,  which  the  equimulti- 
'  pie  of  the  third  has  to  that  of  the  fourth.'* 

Let  A  the  first,  have  to  B  the  second^  the  same  ratio  which  the 
third  C  has  to  the  fourth  D;  and  of  A  and  C  let  there  be  taken 
any  equimultiples  whatever  E,  F^  and  of  B  and  D  any  equimulti- 
ples whatever  G,  H:  then  E  has  the 
same  ratio  to  G,  which  F  has  to  H. 

Take  of  E  and  F  any  equimulti- 
ples whatever  K,  L,  and  of  G,  H, 
any  equimultiples  whatever  M,  N: 
then,  because  E  is  the  same  multiple 
of  A,  that  F  is  of  C;  and  of  E  and 
F  have  been  taken  equimultiples  K, 
L;  therefore  K  is  the  same  multiple 
of  A,  that  L  is  of  C  (3.  5.)?  for  the 
same  reason,  M  is  the  same  multiple 
of  B,  that  N  is  of  D:  and  because, 
as  A  is  to  B,  so  is  C  to  D  (Hypoth.)  K  E 
and  of  A  and  C  have  been  taken  cer- 
tain equimultiples  K,  L;  and  of  B  L  F 
and  D  have  been  taken  certain  equi- 
multiples M,  N;  if  therefore  K  be 
greater  than  M,  L  is  greater  than  N; 
and  if  equal,  equalj  if  less,  less  (5. 
def.  5.).  And  K,  L  are  any  equimul- 
tiples whatever  of  E,  Fj  and  M,  N 
any  whatever  of  G,  H:  as  therefore 
E  is  to  G,  so  is  (5.  def.  5.)  F  to  H. 
Therefore,  if  the  first,  &c,  Q.  E.  D. 

Cor.  Likewise,  if  the  first  have 
the  same  ratio  to  the  second,  which 
the  third  has  to  the  fourth,  then  also 
any  equimultiples  whatever  of  the  first  and  third  have  the  same 
ratio  to  the  second  and  fourth:  and  in  like  manner,  the  first  and 
the  third  have  the  same  ratio  to  any  equimultiples  whatever  of  the 
second  and  fourth. 

Let  A  the  first,  have  to  B  the  second,  the  same  ratio  which  the 
third  C  has  to  the  fourth  D,  and  of  A  and  C  let  E  and  F  be  any 
equimultiples  whatever?  then  E  is  to  B,  as  F  to  D. 

Take  of  E,  F  any  equimultiples  whatever  K,  L,  and  of  B,  D  any 
equimultiples  whatever  G,  H;  then  it  may  be  demonstrated,  as 


B      G     M 
D     H     N 


See  Note. 


102 


THE  ELEMENTS  OF  EUCLID. 


BOOK  V. 


before,  that  K  is  the  same  multiple  of  A,  that  L  is  of  C:  and  be- 
cause A  is  to  B,  as  C  is  to  D,  and  of  A  and  C  certain  equimulti- 
ples have  been  taken,  viz.  K  and  L;  and  of  B  and  D  certain  equi- 
multiples G,  Hj  therefore  if  K  be  greater  than  G,  L  is  greater 
than  H^  and  if  equal,  equal;  if  less,  less  (5.  def.  5.):  and  K,  L  are 
any  equimultiples  of  E,  F,  and  G,  H  any  whatever  of  B,  D;  as 
therefore  E  is  to  B,  so  is  F  to  D:  and  in  the  same  way  the  other 
case  is  demonstrated. 

PROP.  V.  THEOR. 

If  one  magnitude  be  the  same  multiple  of  another, 
which  a  magnitude  taken  from  the  first  is  of  a  magni- 
tude taken  from  the  other;  the  remainder  shall  be  the 
same  multiple  of  the  remainder,  that  the  whole  is  of  the 
whole. 

Let  the  magnitude  AB  be  the  same  multi-         G 
pie  of  CD,  that  AE  taken  from  the  first,  is  of 
CF  taken  from  the  other;  the  remainder  EB 
shall  be  the  same  multiple  of  the  remainder 
FD,  that  the  whole  AB  is  of  the  whole  CD. 

Take  AG  the  same  multiple  of  FD,  that 
AE  is  of  CF:  therefore  AE  is  (l.  5.)  the  same 
multiple  of  CF,  that  EG  is  of  CD;  but  AE,  by 
the  hypothesis,  is  the  same  multiple  of  CF, 
that  AB  is  of  CD,  therefore  EG  is  the  same 
multiple  of  CD  that  AB  is  of  CD;  wherefore 
EG  is  equal  to  AB  (l.  Ax.  5.).  Take  from 
them  the  common  magnitude  AE;  the  re- 
mainder AG  is  equal  to  the  remainder  EB. 
Wherefore,  since  AE  is  the  same  multiple  of 
CF,  that  AG  is  of  FD,  and  that  AG  is  equal 
to  EB;  therefore  AE  is  the  same  multiple  of 
CF,  that  EB  is  of  FD :  but  AE  is  the  same  multiple  of  CF,  that  AB 
is  of  CD;  therefore  EB  is  the  same  multiple  of  FD,  that  AB  is  of 
CD.     Therefore,  if  any  magnitude,  Sec.  Q.  E.  D. 


A— 


E. 


F_ 


D 


PROP.  VI.  THEOR. 

If  two  magnitudes  be  equimultiples  of  two  others, 
and  if  equimultiples  of  these  be  taken  from  the  first  two, 
the  remainders  are  either  equal  to  these  others,  or  equi- 
multiples of  them.* 

Let  the  two  magnitudes  AB,  CD  be  equimultiples  of  the  two 
E,  F,  and  AG,  CH  taken  from  the  first  two  be  equimultiples  of 
the  same  E,  F;  the  remainders  GB,  HD  are  either  equal  to  E,  F, 
or  equimultiples  of  them. 


See  Note. 


ROOK  V. 


THE    ELEMENTS    OF  EUCLID. 


lo; 


G— 


K 


First,  let  GB  be  equal  to  E,  HD  is  A 
equal  to  F:  make  CK  equal  to  F;  and 
because  AG  is  the  same  multiple  of 
E,  that  CH  is  of  F,  and  that  GB  is 
equal  to  E,  and  CK  to  F;  therefore 
AB  is  the  same  multiple  of  E,  that 
KHisofF.  But  AB,  by  the  hypothe- 
sis, is  the  same  multiple  of  E  that  CD 
is  of  F;  therefore  KH  is  the  same  mul- 
tiple of  F,  that  CD  is  of  Y;  wherefore 
KH  is  equal  to  CD  (l.  Ax.  5.):  take 
away  the  common  magnitude  CH, 
then  the  remainder  KC  is  equal  to  the 
remainder  UD;  but  KC  is  equal  to  F;  HD  therefore  is  equal  to  F. 


H— 


B 


D      E     F 


K 


H 


But  let  GB  be  a  multiple  of  E:  then 
HD  is  the  same  multiple  of  F:  make 
CK  the  same  multiple  of  F,  that  GB 
is  of  E:  and  because  AG  is  the  same 
multiple  of  E,  that  GH  is  of  F;  and 
GB  the  same  multiple  of  E  that  CK 
is  of  F:  therefore  AB  is  the  same  mul- 
tiple of  E,  thatKH  is  of  F  (2.  5.):  but 
AB  is  the  same  multiple  of  E,  that  CD 
is  of  F,  therefore  KH  is  the  same  mul- 
tiple of  F,  that  CD  is  of  it:  wherefore 
KH  is  equal  to  CD  (1.  Ax.  5.):  take 
away  CH  from  both:  therefore  the  re- 
mainder KC  is  equal  to  the  remainder 
HD:  and  because  GB  is  the  same  mul- 
tiple of  E,  that  KC  is  of  F,  and  that  KC  is  equal  to  HD,-  therefore 
HD  is  the  same  multiple  of  F,  that  GB  is  of  E.  If  therefore  two 
magnitudes,  Sec.  Q.  E.  D. 


F 


PROP.  A.  THEOR. 

If  the  first  of  four  magnitudes  have  to  the  second  the 
same  ratio  which  the  third  has  to  the  fourth;  then,  if 
the  first  be  greater  than  the  second,  the  third  is  also 
greater  than  the  fourth;  and  if  equal,  equal;   if  less, 

less.* 

Take  any  equimultiples  of  each  of  them,  as  the  doubles  of  each; 
then,  by  def.  5th  of  this  book,  if  the  double  of  the  first  be  greater 
than  the  double  of  the  second,  the  double  of  the  third  is  greater 
than  the  double  of  the  fourth;  but,  if  the  first  be  greater  than  the 
second,  the  double  of  the  first  is  greater  than  the  double  of  the 
second;  wherefore  also  the  double  of  the  third  is  greater  than  the 
double  of  the  fourth:  therefore  the  third  is  greater  than  the 
fourth:  in  like  manner,  if  the  first  be  equal  to  the  second,  or  less 
than  it,  the  third  can  be  proved  to  be  equal  to  the  fourth,  or  less 
than  it.     Therefore,  if  the  first,  &c.  Q.  E.  D. 


*  See  Notes 


104 


THE  ELEMENTS  OF  EUCLID. 


BOOK  V. 


G 

H 


A 
C 


B 
D 


E 
F 


PROP.  B.  THEOR. 

If  four  magnitudes  be  proportionals,  they  are  pro- 
portionals also  when  taken  inversely.^ 

If  the  magnitude  A  be  to  B,  as  C  is  to  D,  then  also  inversely  B 
is  to  A,  as  D  to  C. 

Take  of  B  and  D  any  equimultiples  what- 
ever E  and  F;  and  of  A  and  C  any  equimul- 
tiples whatever  G  and  H.  First,  let  E  be 
greater  than  G,  then  G  is  less  than  E;  and 
because  A  is  to  B,  as  C  is  to  D,  and  of  A  and 

C,  the  first  and  third,  G  and  H  are  equimul- 
tiples; and  of  B  and  D,  the  second  and  fourth, 
E  and  F  are  equimultiples;  and  that  G  is  less 
than  E,  H  is  also  (5.  def.  5.)  less  than  F;  that 
is,  F  is  greater  than  H;  if  therefore  E  be 
greater  than  G,  F  is  greater  than  H;  in  like 
manner,  if  E  be  equal  to  G,  F  may  be  shown 
to  be  equal  to  H;  and,  if  less,  less;  and  E,  F 
are    any  equimultiples  whatever    of   B    and 

D,  and  G,  H  any  whatever  of  A  and  C;  there- 
fore, as  B  is  to  A,  so  is  D  to  C.     If,  then,  four  magnitudes,  &c. 
Q.  E.  D. 

PROP.  C.  THEOR. 

If  the  first  be  the  same  multiple  of  the  second,  or  the 
same  part  of  it,  that  the  third  is  of  the  fourth ;  the  first 
is  to  the  second,  as  the  third  is  to  the  fourth.^ 

Let  the  first  A  be  the  same  multiple  of  B 
the  second,  that  C  the  third  is  of  the  fourth 
D:  A  is  to  B  as  C  is  to  D. 

Take  of  A  and  C  any  equimultiples  what- 
ever E  and  F;  and  of  B  and  D  any  equi- 
multiples whatever  G  and  H;  then,  because 
A  is  the  same  multiple  of  B  that  C  is  of  D; 
and  that  E  is  the  same  multiple  of  A,  that 
F  is  of  C;  E  is  the  same  multiple  of  B  that 
F  is  of  D  (3.  5.);  therefore  E  and  F  are  the 
same  multiples  of  B  and  D:  but  G  and  H 
are  equimultiples  of  B  and  D:  therefore,  if 
E  be  a  greater  multiple  of  B,  than  G  is,  F  is 
a  greater  multiple  of  D,  than  H  is  of  D: 
that  is,  if  E  be  greater  than  G,  F  is  greater 
than  H:  in  like  manner,  if  E  be  equal  to  G, 
or  less;  F  is  equal  to  H,  or  less  than  it. 
But  E,  F  are  any  equimultiples  v/hatever  of 
A,  C,  and  G,  H  any  equimultiples  whatever 
of  B,  D.  Therefore  A  is  to  B,  as  C  is  to  D 
(5.  def.) 

"  See  Notes. 


A 
E 


B 


G 


C 

F 


D 
H 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


105 


Next,  Let  the  first  A  be  the  same  part  of 
the  second  B,  that  the  third  C  is  of  the  fourth 
D:  A  is  to  B,  as  C  is  to  D:  for  B  is  the 
same  multiple  of  A,  that  D  is  of  C:  where- 
fore, by  the  preceding  case,  B  is  to  A,  as  D 
is  to  C;  and  inversely  (B.  5.)  A  is  to  B,  as 
C  is  to  D.  Therefore,  if  the  first  be  the 
same  multiple,  &:c.  Q.  E.  D. 


BCD 


A      B      C      D 


E 


PROP.  D.  THEOR. 

If  the  first  be  to  the  second  as  the  third  to  the 
fourth,  and  if  the  first  be  a  multiple,  or  part  of  the  se- 
cond; the  third  is  the  same  multiple,  or  the  same  part 
of  the  fourth."^ 

Let  A  be  to  B,  as  C  is  to  D;  and  first  let  A  be  a  multiple  of  B, 
C  is  the  same  multiple  of  D. 

Take  E  equal  to  A,  and  whatever  multi- 
ple A  or  E  is  of  B,  make  F  the  same  multi- 
ple of  D:  then,  because  A  is  to  B,  as  C  is  to 
D;  and  of  B  the  second,  and  D  the  fourth 
equimultiples  have  been  taken  E  and  F;  A 
is  to  E,  as  C  to  F  (Cor.  4.  5.):  but  A  is  equal 
to  E,  therefore  C  is  equal  to  F  (A.  5.):  and 
F  is  the  same  multiple  of  D,  that  A  is  of  B. 
Wherefore  C  is  the  same  multiple  of  D,  that 
A  is  of  B. 

Next,  Let  the  first  A  be  a  pa4't  of  the  se- 
cond B;  C  the  third  is  the  same  part  of  the 
fourth!  D. 

Because  A  is  to  B,  as  C  is  to  D;  then  in- 
versely, B  is  (B.  5.)  to  A,  as  D  to  C:  but  A 
is  a  part  of  B,  therefore  B  is  a  multiple  of  A; 
and,  by  the  preceding  case,  D  is  the  same 
multiple  of  C,  that  is,  C  is  the  same  part  of 
D,  that  A  is  of  B;  therefore,  if  the  first.  Sec. 
Q.  E.  D. 

PROP.  VIL  THEOR. 

Equal  magnitudes  have  the  same  ratio  to  the  same 
magnitude;  and  the  same  has  the  same  ratio  to  equal 
magnitudes.  , 

Let  A  and  B  be  equal  magnitudes,  and  C  any  other.  A  and  B 
have  each  of  them  the  same  ratio  to  C;  and  C  has  the  same  ratio 
to  each  of  the  magnitudes  A  and  B. 

Take  of  A  and  B  any  equimultiples  whatever  D  and  E,  and  of 
C  any  multiple  whatever  F;  then,  because  D  is  the  same  multi- 


O 


*  See  Note 


See  the  figure  above. 


106 


THE  ELEMENTS  OF  EUCLID. 


BOOK  V. 


pie  of  Aj  that  E  is  of  B,  and  that  A  is  equal 
to  B;  D  is  (l.  Ax.  5.)  equal  to  E:  therefore, 
if  D  be  greater  than  F,  E  is  greater  than  F: 
and  if  equal,  equals  if  less,  less:  andD,Eare 
any  equiniultiples  of  A,  B,  and  F  is  any  mul- 
tiple of  C.     Therefore  (5.  def.  5.)  as  A  is  to 

C,  so  is  B  to  C. 

Likewise  C  has  the  same  ratio  to  A,  that 
it  has  to  B:  for  having  made  the  same  con- 
struction, D  mav  in  like  manner  be  shown 
equal  to  E:  therefore,  if  F  be  greater  than 

D,  it  is  likewise  greater  than  E;  and  if  equal, 
equal;  if  less,  less:  and  F  is  any  multiple 
whatever  of  C,  and  D,  E  are  any  equimulti- 
ples whatever  of  A,  B.  Therefore,  C  is  to 
A,  as  C  is  to  B  (5.  def.  5.).  Therefore  equal 
magnitudes,  Sec.  Q.  E.  D. 


D 

E 


A 
B 


PROP.  VIII.  THEOR. 

Of  unequal  magnitudes,  the  greater  has  a  greater  ra« 
tio  to  the  same  than  the  less  has;  and  the  same  mag- 
nitude has  a  greater  ratio  to  the  less,  than  it  has  to  the 
greater.^ 

Let   AB,  BC  be   unequal   magnitudes,  of  which   AB  is  the 


E 


F— 


Fig.  1. 


C_ 


greater,  and  let  D  be  any  magnitude  what- 
ever: AB  has  a  greater  ratio  to  D,  than 
BC  to  D;  and  D  has  a  greater  ratio  to 
BC  than  to  AB. 

If  the  magnitude  which  is  not  the 
greater  of  the  two  AC,  CB,  be  not  less 
than  D,  take  EF,  FG,  the  doubles  of 
AC,  CB,  as  in  Fig.  1 .  But,  if  that  which 
is  not  the  greater  of  the  two  AC,  CB  be 
less  than  D  (as  in  Fig.  2.  and  3.)  this 
magnitude  can  be  multiplied,  so  as  to 
become  greater  than  D,  whether  it  be 
AC,  or  CB.  Let  it  be  multiplied,  until 
it  become  greater  than  D;  and  let  the 
other  be  multiplied  as  often;  and  let  EF 
be  the  multiple  thus  taken  of  AC,  and 
FG  the  same  multiple  of  CB;  therefore 
EF  and  FG  are  each  of  them  greater  than 
D:  and  in  every  one  of  the  cases,  take  H  the  double  of  D,  K,  its 
triple,  and  so  on,  till  the  multiple  of  D  be  that  which  first  be- 
comes greater  than  FG:  let  L  be  that  multiple  of  D  which  is 
first  greater  than  FG,  and  K  the  multiple  of  D  which  is  next  less 
than  L. 


G 


K     H 


D 
I 


Seo  Note. 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


107 


F— 


C— 


F—C- 


Then,  because  L  is  the  multiple  of  D,  which  is  the  first  that 
becomes  greater  than  FG;  the  next  preceding-  multiple  K  is  not 
greater  than  FG;  that  is,  FG  is  not  less  than  K:  and  since  EF  is 
the  same  multiple  of  AC,  that  FG  is  of  CB;  FG  is  the  same  mul- 
tiple of  CB,  that  EG  is  of  AB  (l.  5.):  wherefore  EG  and  FG  are 
equimultiples  of  AB 

and  CB:   and   it  was  Fig.  2.  Fig.  3. 

shown,  that  FG  was  E 
not  less  than  K,  and, 
by  the  construction, 
EF  is  greater  than  Dj 
therefore  the  whole 
EG  is  greater  than  K 
and  D  together;  but, 
K  together  with  D,  is 
equal  to  h;  therefore 
EG  is  greater  than  L; 
but  FG  is  not  greater 
than  L;  and  EG,  FG 
are  equimultiples  of 
AB,  BC,  and  L  is  a 
multiple  of  D;  there- 
fore (7.  def.  5.)  AB 
has  to  D  a  greater  ra- 
tio than  BC  has  to  D. 

Also  D  has  to  BC  a 
greater  ratio  than  it 
has  to  AB,  for,  hav- 
ing made  the  same 
construction,  it  may 
be  shown,  in  like  man- 
ner, that  L  is  greater 
than  FG,  but  that  it  is 
not  greater  than  EG:  and  L  is  a  multiple  of  D;  and  FG,  EG  are 
equimultiples  of  CB,  AB;  therefore  D  has  to  CB  a  greater  ratio 
(7.  def.  5.)  than  it  has  to  AB.  Wherefore,  of  unequal  magni- 
tudes, Sec.  Q.  E.  D. 


B 


L     K   H   D 


G     B 

L     K     D 


PROP.  IX.  THEOR. 

Magnitudes  which  have  the  same  ratio  to  the  same 
magnitude  are  equal  to  one  another;  and  those  to 
which  the  same  magnitude  has  the  same  ratio  are  equal 
to  one  another."* 

Let  A,  B  have  each  of  them  the  same  ratio  to  C:  A  is  equal  to 
B:  for  if  they  be  not  equal,  one  of  them  is  greater  than  the  other; 
let  A  be  the  greater;  then,  by  what  was  shown  in  the  preceding 


*  See  Note. 


108 


THE  ELEMENTS  OF  EUCLID. 


BOOK  V. 


proposition,  there  are  some  equimultiples  of  A  and  B,  and  some 
multiple  of  C  such,  that  the  multiple  of  A  is  greater  than  the  mul- 
tiple of  C,  but  the  multiple  of  B  is  not  greater  than  that  of  C. 
Let  such  multiples  be  taken,  and  let  D,  E,  be  the  equimultiples  of 
A,  B,  and  F  the  multiple  of  C,  so  that  D  may  be  greater  than  F, 
and  E  not  greater  than  F:  but,  because  A  is  to  C,  as  B  is  to  C, 
and  of  A,  B  are  taken  equimultiples 
D,  E,  and  of  C  is  taken  a  multiple  F; 
and  that  D  is  greater  than  F;  E  shall 
also  be  greater  than  F  (5.  def.  5.):  but  -^ 

E  is  not  greater  than  F,  which  is  im-      A 
possible;  A  therefore  and  B  are  not 
unequal;  that  is,  they  are  equal. 

Next,  let  C  have  the  same  ratio  to 
each  of  the  magnitudes  A  and  B;  A 
is  equal  to  B:  for  if  they  be  not,  one  E 

of  them  is  greater  than  the  other;  let  B 
A  be  the  greater;  therefore,  as  was 
shown  in  Prop.  8th,  there  is  some 
multiple  F  of  C,  and  some  equimulti- 
ples E  and  D,  of  B  and  A  such,  that  F  is  greater  than  E,  and  not 
greater  than  D;  but  because  C  is  to  B,  as  C  is  to  A,  and  that  F, 
the  multiple  of  the  first,  is  greater  than  E,  the  multiple  of  the  se- 
cond; F  the  multiple  of  the  third,  is  greater  than  D,  the  multiple 
of  the  fourth  (5.  def.  5.):  but  F  is  not  greater  than  D,  which  is 
impossible.  Therefore  A  is  equal  to  B.  Wherefore,  magnitudes 
which,  8cc.  Q.  E.  D. 


PROP.  X.  THEOR. 

That  magnitude  which  has  a  greater  ratio  than  ano- 
ther has  unto  the  same  magnitude,  is  the  greater  of  the 
two:  and  that  magnitude,  to  which  the  same  has  a 
greater  ratio  than  it  has  unto  another  magnitude  is 
the  lesser  of  the  two.^ 

Let  A  have  to  C  a  greater  ratio  than  B  has  to  C:  A  is  greater 
than  B:  for,  because  A  has  a  greater  ratio  to  C,  than  B  has  to  C, 
there  are  (7.  def.  5.)  some  equimultiples  of  A  and  B,  and  some 
multiple  of  C  such,  that  the  multiple  of  A  is  greater  than  the  mul- 
tiple of  C,  hut  the  multiple  of  B,  is  not  greater  than  it:  let  them 


See  Note. 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


109 


be  taken,  and  let  D,  E  be  equimulti- 
ples of  A,  B,  and  F  a  multiple  of  C 
such,  that  D  is  greater  than  F,  but  E 
is  not  greater  than  F:  therefore  D  is  ^ 
greater  than  E:  and,  because  D  and 
E  are  equimultiples  of  A  and  B,  and 
D  is  greater  than  E;  therefore  A  is 
(4.  Ax.  5.)  greater  than  B. 

Next,  let  C  have  a  greater  ratio  to 
B  than  it  has  to  Aj  B  is  less  than  A: 
for  there  is  some  multiple  F  of  C,  and 
some  equimultiples  E  and  D  of  B  and 
A  such,  that  F  is  greater  than  E,  but 
it  is  not  greater  than  D:  E  therefore  g 
is  less  than  D;  and  because  E  and  D 
are  equimultiples  of  B  and  A,  there- 
fore B  is  (4.  Ax.  5.)  less  than  A. 
That  magnitude,  therefore,  &c.  Q. 
E.  D. 


D 


E 


PROP.  XI.  THEOR. 

Ratios  that  are  the  same  to  the  same  ratio,  are  the 
same  to  one  another. 

Let  A  be  to  B  as  C  is  to  D,*  and  as  C  to  D,  so  let  E  be  to  F;  A 
is  to  B  as  E  to  F. 

Take  of  A,  C,  E  any  equimultiples  whatever  G,  H,  K;  and  of 
B,  D,  F  any  equimultiples  whatever  L,  M,  N.  Therefore,  since 
A  is  to  B,  as  C  to  D,  and  G,  H  are  taken  equimultiples  of  A,  C, 
and  L,  M  of  B,  D,*  if  G  be  greater  than  L,  H  is  greater  than  M; 
and  if  equal,  equals  and  if  less,  less  (5.  def.  5.).  Again,  because 
C  is  to  D,  as  E  to  F,  and  H,  K  are  taken  equimultiples  of  C,  E: 
and  M,  N,  of  D,  F:  if  H  be  greater  than  M,  K  is  greater  than  N^ 
and  if  equal,  equal j  and  if  less,  less:  but  if  G  be  greater  than  L, 


H- 


K- 


A- 
B- 
L- 


C- 

D- 

M- 


E- 
F- 

N- 


it  has  been  shown  that  H  is  greater  than  M;  and  if  equal,  equal; 
and  if  less,  less:  therefore,  if  G  be  greater  than  L,  K  is  greater 
than  N;  and  if  equal,  equal;  and  if  less,  less:  and  G,  K  are  any 
equimultiples  whatever  of  A,  E;  and  L,  N  any  whatever  of  B,  F: 
therefore  as  A  is  to  B,  so  is  E  to  F  (5.  def.  5.).  Wherefore,  ratios 
that,  &c.  Q.  E.  D. 


1  10  THE  ELEMENTS  OF  EUCLID.  BOOK  V. 

PROP.  XII.  THEOR. 

If  any  number  of  magnitudes  be  proportionals  as 
one  of  the  antecedents  is  to  its  consequent,  so  shall 
all  the  antecedents  taken  together  be  to  all  the  conse- 
quents. 

Let  any  number  of  magnitudes  A,  B,  C,  D,  E,  F  be  proportion- 
als; that  is,  as  A  is  to  B,  so  is  C  to  D,  and  E  to  F:  as  A  is  to  B, 
so  shall  A,  C,  E  together  be  to  B,  D,  F  together. 

Take  of  A,  C,  E  any  equimultiples  whatever  G,  H,  K;  and  of 


G H K- 

A C E- 

B D F- 


L M ^ N 

B,  D,  F  any  equimultiples  whatever  L,  M,  N:  then,  because  A  is 
to  B,  as  C  is  to  D,  and  E  to  F;  and  that  G,  H,  K  are  equimulti- 
ples of  A,  C,  E,  and  L,  M,  N  equimultiples  of  B,  D,  F5  if  G  be 
greater  than  L,  H  is  greater  than  M,  and  K  greater  than  N;  and 
if  equal,  equal;  and  if  less,  less  (5.  def.  5.).  Wherefore,  if  G  be 
greater  than  L,  then  G,  H,  K  together  are  greater  than  L,  M,  N, 
together;  and  if  equal,  equal;  and  if  less,  less.  And  G,  and  G, 
H,  K,  together  are  any  equimultiples  of  A,  and  A,  C,  E  together; 
because,  if  there  be  any  number  of  magnitudes  equimultiples  of 
as  many,  each  of  each,  whatever  multiple  one  of  them  is  of  its 
part,  the  same  multiple  is  the  whole  of  the  whole  (1.  5.):  for  the 
same  reason  L,  and  L,  M,  N  are  any  equimultiples  of  B,  and  B, 
D,  F:  as  therefore  A  is  to  B,  so  are  A,  C,  E  together  to  B,  D,  F 
together.     Wherefore,  if  any  number,  &c.  Q.  E.  D. 

PROP.  XIII.  THEOR. 

If  the  first  has  to  the  second  the  same  ratio  which 
the  third  has  to  the  fourth,  but  the  third  to  the  fourth  a 
greater  ratio  than  the  fifth  has  to  the  sixth;  the  first, 
shall  also  have  to  the  second  a  greater  ratio  than  the 
fifth  has  to  the  sixth."^ 

Let  A  the  first  have  the  same  ratio  to  B  the  second,  which  C 
the  third,  has  to  D  the  fourth,  but  C  the  third  to  D  the  fourth,  a 
greater  ratio  than  E  the  fifth  to  F  the  sixth:  also  the  first  A 
shall  have  to  the  second  B  a  greater  ratio  than  the  fifth  E  to  the 
sixth  F. 

Because  C  has  a  greater  ratio  to  D,  than  E  to  F,  there  are  some 
equimultiples  of  C  and  F,  and  some  of  D  and  F  such,  than  the 
multiple  of  C  is  greater  than  the  multiple  of  D,  but  the  multiple 

""  See  Note,  " 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


Ill 


M- 
A- 
B- 

N- 


G- 


H- 
E- 


D- 
K- 


F- 
L- 


of  E  is  not  greater  than  the  multiple  of  F  (7.  def.  5.),*  let  such  be 
taken^  and  of  C,  E  let  G,  H  be  equimultiples,  and  K,  L  equimul- 
tiples of  D,  F,  so  that  G  be  greater  than  K,  but  H  not  greater 
than  L;  and  whatever  multiple  G  is  of  C,  take  M  the  same  multi- 
ple of  A;  and  whatever  multiple  K  is  of  D,  take  N  the  same  mul- 
tiple of  B;  then,  because  A  is  to  B,  as  C  to  D,  and  of  A  and  C, 
M  and  G  are  equimultiples:  and  of  B  and  D,  N  and  K  are  equi- 
multiples; if  M  be  greater  than  N,  G  is  greater  than  K,  and  if 
equal,  equal;  and  if  less,  less  (5.  def.  5.):  but  G  is  greater  than 
K,  therefore  M  is  greater  than  N:  but  H  is  not  greater  than  L; 
and  M,  H  are  equimultiples  of  A,  E;  and  N,  L  equimultiples  of 
B,  F:  therefore  A  has  a  greater  ratio  to  B,  than  E  has  to  F  (7. 
def.  5.).     Wherefore,  if  the  first,  &c.  Q.  E.  D. 

Cor.  And  if  the  first  have  a  greater  ratio  to  the  second,  than 
the  third  has  to  the  fourth,  but  the  third  the  same  ratio  to  the 
fourth,  which  the  fifth  has  to  the  sixth:  it  may  be  demonstrated, 
in  like  manner,  that  the  first  has  a  greater  ratio  to  the  second, 
than  the  fifth  has  to  the  sixth. 

PROP.  XIV.  THEOR. 

If  the  first  has  to  the  second,  the  same  ratio  which 
the  third  has  to  the  fourth;  then,  if  the  first  be  greater 
than  the  third,  the  second  shall  be  greater  than  the 
fourth;  and  if  equal,  equal;  and  if  less,  less.* 

Let  the  first  A  have  to  the  second  B,  the  same  ratio  which  the 
third  C  has  to  the  fourth  D;  if  A  be  greater  than  C,  B  is  greater 
than  D. 

Because  A  is  greater  than  C,  and  B  is  any  other  magnitude,  A 
has  to  B  a  greater  ratio  than  C  to  B  (8.  5.);  but  as  A  is  to  B,  so 


A     B     C      D 


BCD 


CD         A 

is  C  to  D;  therefore  also  C  has  to  D  a  greater  ratio  than  C  has  to 
B  (13.  5.):  but  of  two  magnitudes,  that  to  which  the  same  has 
the  greater  ratio  is  the  lesser  (10.  5.):  wherefore  D  is  less  than 
B;  that  is,  B  is  greater  than  D. 

Secondly,  If  A  be  equal  to  C,  B  is  equal  to  D;  for  A  is  to  B, 
as  C,  that  is  A,  to  D;  B  therefore  is  equal  to  D  (9.  5.). 

*  See  Note. 


K- 
L~ 


112  THE  ELEMENTS  OF  EUCLID.  BOOK  V« 

Thirdly,  If  A  be  less  than  C,  B  shall  be  less  than  D;  for  C  is 
greater  than  A,  and  because  C  is  to  D,  as  A  is  to  B,  D  is  greater 
than  B,  by  the  first  case;  wherefore  B  is  less  than  D.  Therefore, 
if  the  first,  &c.  Q.  E.  D. 

PROP.  XV.  THEOR. 

Magnitudes  have  the  same  ratio  to  one  another 
which  their  equimultiples  have. 

Let  AB  be  the  same  multiple  of  C,  that  DE  i^  of  F5  C  is  to  F, 
as  AB  to  DE. 

Because  AB  is  the  same  multiple  of  C,  that  DE  is  of  F; 
there  are  as  many  magnitudes  in  AB  equal  A 

to  C,  as  there  are  in  DE  equal  to  F:  let 
AB  be  divided  into  magnitudes,  each 
equal  to  C,  viz.  AG,  GH^  HB;  and  DE  ^ 

into  magnitudes,  each  equal  to  F,  viz.  p 
DK,  KL,  LE:  then  the  number  of  the 
first  AG,  GH,  HB,  shall  be  equal  to  the 
number  of  the  last  DK,  KL,  LE:  and 
because  AG,  GH,  HB  are  all  equal,  and  H- 
that  DK,  KL,  LE  are  also  equal  to  one 
another;  therefore  AG  is  to  DK,  as  GH 
to  KL,  and  as  HB  to  LE  (7.  5.):  and 
as  one  of  the  antecedents  to  its  conse- 
quent, so  are  all  the  antecedents  together  to  all  the  consequents 
together  (12.  5.);  wherefore,  as  AG  is  to  DK,  so  is  AB  to  DE: 
but  AG  is  equal  to  C,  and  DK  to  F:  therefore,  as  C  is  to  F,  so  is 
AB  to  DE.     Therefore  magnitudes,  Sec.  Q.  E.  D. 

PROP.  XVL  THEOR. 

If  four  magnitudes  of  the  same  kind  be  proportionals, 
they  shall  also  be  proportionals  when  taken  alternately. 

Let  the  four  magnitudes.  A,  B,  C,  D,  be  proportionals,  viz.  as 
A  to  B,  so  C  to  D:  they  shall  also  be  proportionals,  when  taken 
alternately;  that  is,  A  is  to  C,  as  B  to  D. 

Take  of  A  and  B  any  equimultiples  whatever  E  and  F:  and  of 
C  and  D  take  any  equimultiples  whatever  G  and  H:  and  because 
E  is  the  same  multiple  of  A,  that  F  is  of  B,  and  that  magnitudes 
have  the  same  ratio  to  one  another  which  their  equimultiples 
have  (15.  5.);  therefore  A  is  to  B,  as  E  is  to  F:  but  as  A  is  to  B, 
so  is  C  to  D:  wherefore 

as  C  is  to  D,  so  (1 1.  5.)  is     E G 

E   to  F:    again,   because 

G,  H    are   equimultiples     A C 

of  C,  D,  as  C  is  to  D,  so 

is  G  to  H  (15.  5.);  but  as     B D^^ 

C  is  to  D,  so  is  E  to  F. 

Wherefore  as  E  is  to  F,     F H 

so   is   G   to   H  (11.    5.). 

But,  when   four   magnitudes  are  proportionals,   if  the   first  be 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


113 


greater  than  the  third,  the  second  shall  be  greater  than  the  fourth; 
and  if  equal,  equal;  if  less,  less  (14.  5.).  Wherefore,  if  E  be 
greater  than  G,  F  likewise  is  greater  than  H;  and  if  equal,  equal; 
if  less,  less;  and  E,  F  are  any  equimultiples  whatever  of  A,  B; 
and  G,  H  any  whatever  of  C,  D.  Therefore  A  is  to  C,  as  B  to  D 
(5.  def.  5.).     If  then  four  magnitudes,  Sec.  Q.  E.  D. 

PROP.  XVII.  THEOR. 

If  magnitudes,  taken  jointly,  be  proportionals,  they 
shall  also  be  proportionals  when  taken  separately;  that 
is,  if  two  magnitudes  together  have  to  one  of  them  the 
same  ratio  which  two  others  have  to  one  of  these,  the 
remaining  one  of  the  first  two  shall  have  to  the  other 
the  same  ratio  which  the  remaining  one  of  the  last  two 
has  to  the  other  of  these."^ 

Let  AB,  BE,  CD,  DF  be  the  magnitudes  taken  jointly  which 
are  proportionals;  that  is,  as  AB  to  BE,  so  is  CD  to  DF;  they 
shall  also  be  proportionals  taken  separately,  viz.  as  AE  to  EB,  so 
CF  to  FD. 

Take  of  AE,  EB,  CF,  FD,  any  equimultiples  whatever  GH, 
HK,  LM,  MN;  and  again,  of  EB,  FD,  take  any  equimultiples 
whatever  KX,  NP:  and  because  GH  is  the  same  multiple  of  AE, 
that  HK  is  of  EB,  therefore  GH  is  the  same  multiple  (1.  5.)  of 
AE,  that  GK  is  of  AB;  but  GH  is  the  same  multiple  of  AE,  that 
LM  is  of  CF:  wherefore  GK  is  the  same  multiple  of  AB,  that 
LM  is  of  CF.  Again,  because  LM  is  the  same  multiple  of  CF, 
that  MN  is  of  FD;  therefore  LM  is  the  same  multiple  (1.  5.)  of 
CF,  that  LN  is  of  CD:  but  LM  v/as  shown  to  be  the  same  mul- 
tiple of  CF,  that  GK  is  of  AB;  GK  therefore  is  the  same  mul- 
tiple of  AB,  that  LN  is  of  CD;  that  is,  GK,  LN  are  equimulti- 
ples of  AB,  CD.  Next,  because  HK  is  the  samie  multiple  of  EB, 
that  MN  is  of  FD;  and  that  KX  is  also 
the  same  multiple  of  EB,  that  NP  is 
of  FD;  therefore  HX  is  the  same  mul- 
tiple (2.  5.)  of  EB,  that  MP  is  of  FD. 
And  because  AB  is  to  BE,  as  CD  is 
to  DF,  and  that  of  AB  and  CD,  GK 
and  LN  are  equimultiples,  and  of  EB 
and  FD,  HX  and  MP  are  equimulti- 
ples; if  GK  be  greater  than  HX,  then 
LN  is  greater  than  MP;  and  if  equal, 
equal;  and  if  less,  less  (5.  def.  5.):  but 
if  GH  be  greater  than  KX,  by  adding 
the  common  part  HK  to  both,  GK  is 
greater  than  HX;  wherefore  also  LN 
is  greater  than  MP;  and  by  taking 
away  MN  from  both,  LM  is  greater 
than  NP:  therefore,  if  GH  be  greater 


X 


K— 


H— 


N— 


M— 


B 


D 


E—      F— 


G 


*  See  Note. 


114 


THE  ELEMENTS  OF    EUCLID. 


BOOK  V. 


than  KX,  LM  is  greater  than  NP.  In  like  manner  it  may  be 
demonstrated,  that  if  GH  be  equal  to  KX,  LN  likewise. is  equaL 
to  NP;  and  if  less,  less:  and  GH,  LM  are  any  equimultiples  whal-' 
ever  of  AE,  CF,  and  KX,  NP  are  any  whatever  oiEB,  FD.  There- 
fore (5.  def.  5.),  as  AE  is  to  EB,  so  is  CF  to  FD.  If  then  mag- 
nitudes, &c.  Q.  E.  D. 

PROP.  XVIII.  THEOR. 


If  magnitudes,  taken  separately,  be  proportionals, 
they  shall  also  be  proportionals  when  taken  jointly, 
that  is,  if  the  first  be  to  the  second,  as  the  third  to  the 
fourth,  the  first  and  second  together  shall  be  to  the  se- 
cond, as  the  third  and  fourth  together  to  the  fourth."* 

Let  AE,  EB,  CF,  FD  be  proportionals;  that  is,  as  AE  to  EB, 
so  is  CF  to  FD;  they  shall  also  be  proportionals  when  taken 
jointly;  that  is,  as  AB  to  BE,  so  CD  to  DF. 

Take  of  AB,  BE,  CD,  DF  any  equimultiples  whatever  GH, 
HK,  LM,  MN;  and  again,  of  BE,  DF,  take  any  whatever  equi- 
multiples KO,  NP:  and  because  KO,  NP  are  equimultiples  of 
BE,  DF;  and  that  KH,  NM  are  equimultiples  likewise  of  BE, 
DF,  if  KO,  the  multiple  of  BE,  be  greater  than  KH,  which  is  a 
multiple  of  the  same  BE,  NP,  likewise  the  multiple  of  DF,  shall 
be  greater  than  NM,the  multiple  of 
the  same  DF:  and  if  KO  be  equal  to 
KH,  NP  shall  be  equal  to  NM;  and 
if  less,  less. 

First,  Let  KO  not  be  greater  than 
KH,  therefore  NP  is  not  greater  than 
NM;  and  because  GH,  HK  are  equi- 
multiples of  AB,  BE,  and  that  AB 
is  greater  than  BE,  therefore  GH 
is  greater  (3.  Ax.  5.)  than  KH:  but 
KO  is  not  greater  than  KH,  where- 
fore GH  is  greater  than  KO.  In  like 
manner  it  may  be  shown,  that  LM  _ 

is  greater  than  NP.  Therefore  if  KO  -c,  I        t:,  I 

be  not  greater  than  KH,  then  GH, 
the  multiple  of  AB,  is  always  greater 
than  KO,  the  multiple  of  BE;  and 
likewise  LM,  the  multiple  of  CD, 
greater  than  NP,  the  multiple  of  DF. 

Next,  Let  KO  be  greater  than  KH:  therefore,  as  has  been 
shov/n,  NP  is  greater  that  NM:  and  because  the  whole  GH  is 
the  same  multiple  of  the  whole  AB,  that  HK  is  of  BE,  the  re- 


H 
O. 


K— 


M 

pI 

N. 


B 


D 


*  See  Note. 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


115 


o 

H- 


K- 


G 


M— 


N 


B 


D 

fI 


mainder  GK  is  the  same  multiple  of 
'the  remainder  AE  that  GH  is  of  AB 

(5.  5.):  which  is  the  same  that  LM 

is  of  CD.     In  like  'manner,  because 

hM  is  the  same  multiple  of  CD,  that 

MN,  is  of  DF,  the  remainder  LN  is 

the  same  multiple  of  the  remainder 

CF,  that  thp  whole  LM  is  of  the 

whole  CD  (5.  5.):  but  it  was  shown 

that  LM  is  the  same  multiple  of  CD, 

that  GK  is  of  AE;  therefore  GK  is 

the  same  multiple  of  AE,  that  LN 

is  of  CF;  that  is,  GK,  LN  are  equi- 
multiples of  AE,  CF:  and  because 

KO,  NP  are  equimultiples  of  BE, 

DF,  if  from  KO,  NP  there  be  taken 
KH,   NM,   which  are   likewise    equimultiples   of  BE,   DF,  the 
remainders  HO,  MP  are  either  equal  to  BE,  DF,  or  equimulti- 
ples of  them  (6.  5.)-     First,  let  HO,  MP,  be  equal  to  BE,  DF; 
and  because  AE  is  to  EB,  as  CF  to  FD,  and  that  GK,  LN  are 
equimultiples  of  AE,  CF;  GK  shall  be  to  EB,  as  LN  to  FD  (Cor. 
4.  5.):  but  HO  is  equal  to  EB,  and  MH  to  FD;  wherefore  GK  is 
to  HO,  as  LN  to  MP.     If  therefore  GK  be  greater  than  HO,  LN 
is  greater  than  MP;  and  if  equal,  equal;  and  if  less  (Ax.  5.),  less. 
But  let  HO,  MP  be  equimultiples  of  EB,  FD;  and  because  AE 
is  to  EB,  as  CF  to  FD,  and  that  of  AE,  CF  are  taken  equimul- 
tiples GK,  LN,  and  of  EB,  FD,  the  equimultiples  HO,  MP;  if  GK 
be  greater  than   HO,  LN  is  greater     q 
than  MP;  and  if  equal,  equal;  and  if 
less,  less  (5.  def.  5.):  which  was  like- 
wise shown  in  the  preceding  case.     If 
therefore    GH  be   greater  than   KO, 
taking  KH  from  both,  GK  is  greater 
than  HO;  wherefore  also  LN  is  great- 
er than  MP;  and,  consequently,  adding 
NM  to  both,  LM  is  greater  than  NP: 
therefore,  if  GH  be  greater  than  KO, 
LM  is  greater  than  NP.    In  like  man- 
ner it  may  be  shown,  that  if  GH  be 
equal  to  KO,  LM  is  equal  to  NP;  and 
if  less,  less.     And  in  the  case  in  which 
KO  is  not  greater  than  KH,  it  has  been 
shown  that  GH  is  always  greater  than  KO,  and  likewise  LM  than 
NP:  but  GH,  LM  are  any  equimultiples  of  AB,  CD,  and  KO,  NP 
are  any  whatever  of  BE,  DF:  therefore  (5.  def.  5.),  as  AB  is  to 
BE,  so  is  CD  to  DF.     If  then  magnitudes,  &c.  Q.  E.  D. 

PROP.  XIX.  THEOR. 

If  a  whole  magnitude  be  to  a  whole,  as  a  magnitude 
taken  from  the  first,  is  to  a  magnitude  taken  from  the 


H- 


K- 


M- 


N- 


B 

E. 


G| 


■t 


116 


THE  ELEMENTS  OF  EUCLID. 


BOOK  V. 


alternately 

A 


C 

I 


B 


D 


Other;  the  remainder  shall  be  to  the  remainder,  as  the 
whole  to  the  whole. "^ 

Let  the  whole  AB  be  to  the  whole  CD,  as  AE,  a  magnitude 
taken  from  AB,  to  CF,  a  magnitude  taken  from  CD;  the  remain- 
der EB  shall  be  to  the  remainder  FD,  as  the  whole  AB  to  the 
whole  CD. 

Because  AB  is  to  CD,  as  AE  to  CF,  likewise, 
(IC.  5.)  BA  is  to  AE,  as  DC  to  CF;  and  because,  if 
magnitudes,  take  jointly,  be  proportionals,  they  are 
also  proportionals  (17.  5.)  when  taken  separately; 
therefore,  as  BE  is  to  DF,  so  is  EA  to  FC;  and  al- 
ternately, as  BE  is  to  EA,  so  is  DF  to  FC:  but,  as 
AE  to  CF,  so,  by  the  hypothesis,  is  AB  to  CD; 
therefore  also  BE,  the  remainder,  shall  be  to  the 
remainder  DF,  as  the  whole  AB  to  the  whole  CD: 
Wherefore,  if  the  whole,  he.  Q.  E.  D. 

Cor.  If  the  whole  be  to  the  whole,  as  a  magnitude  taken  from 
the  first,  is  to  a  magnitude  taken  from  the  other;  the  remainder 
likewise  is  to  the  remainder,  as  the  magnitude  taken  from  the 
first  to  that  taken  from  the  other:  the  demonstration  is  contained 
in  the  preceding. 

PROP.  E.  THEOR. 

If  four  magnitudes  be  proportionals,  they  are  also 
proportionals  by  conversion,  that  is,  the  first  is  to  its 
excess  above  the  second,  as  the  third  to  its  excess  above 
the  fourth. 

Let  AB  be  to  BE,  as  CD  to  DF;  then  BA  is  to 
AE,  as  DC  to  CF. 

Because  AB  is  to  BE,  as  CD  to  DF,  by  divi- 
sion (17.  5.),  AE  is  to  EB,  as  CF  to  FD,  and  by 
inversion  (B.  5.),  BE  is  to  EA,  as  DF  to  FC. 
Wherefore,  by  composition  (18.  5.),  BA  is  to  AE, 
as  DC  is  to  CF.     If,  therefore,  four,  Sec.  Q.  E.  D. 


E" 


B 


D 


PROP.  XX.  THEOR. 

If  there  be  three  magnitudes,  and  other  three,  which, 
taken  two  and  two,  have  the  same  ratio;  if  the  first  be 
greater  than  the  third,  the  fourth  shall  be  greater  than 
the  sixth;  and  if  equal,  equal;  and  if  less,  less.^ 

Let  A,  B,  C  be  three  magnitudes,  and  D,  E,  F  other  three, 
which,  taken  two  and  two,  have  the  same  ratio,  viz.  as  A  is  to  B, 


See  NotCo 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


iir 


SO  isI>to  E;  and  asB  toC,so  is  E  to  F.  If  A  be 
greater  than  C,  D  shall  be  greater  than  F;  and  if 
equal,  equal;  and  if  less,  less. 

Because  A  is  greater  than  C,  and  B  is  any  other 
magnitude,  and  that  the  greater  has  to  the  same 
magnitude  a  greater  ratio  than  the  less  has  to  it 
(8.  5,);  therefore  A  has  to  B  a  greater  ratio  than 
C  has  to  B;  but  as  D  is  to  E,  so  is  A  to  B;  there- 
fore (13.  5.)  D  has  to  E  a  greater  ratio  than  C  to     A       B     C 
B;  and  because  B  is  to  C,  as  E  to  F,  by  inversion,     D       E      F 
C  is  to  B,  as  F  is  to  E;  and  D  was  shown  to  have 
to  E  a  greater  ratio  than  C  to  B;  therefore  D  has 
to  E  a  greater  ratio  than  F  to  E  (Cor.  13.  5.);  but 
the  magnitude  which  has  a  greater  ratio  than  an- 
other to  the  same  magnitude,  is  the  greater  of  the  two  (10.  5.);  D 
is  therefore  greater  than  F. 

Secondly,  Let  A  be  equal  to  C;  D  shall  be  equal  to  F:  because 
A  and  C  are  equal  to  one  another, 
A  is  to  B  as  C  is  to  B  (7.  5.):  but  A 
is  to  B,  as  D  to  E;  and  C  is  to  B,  as 
F  to  E:  wherefore  D  is  to  E^  as  F  to 
E  (11.  5.);  and  therefore  D  is  equal 
to  F  (9.  5.).  ABC  ABC 

Next,  let  A  be  less   than  C;DDEF  DEF 

shall  be  less  than  F:  for  C  is  greater 
than  A,  and  as  was  shown  in  the  first 
case,  C  is  to  B,  as  F  to  E,  and  in  like 
manner  B  is  to  A,  as  E  to  D;  there- 
fore F  is  greater  than  D,  by  the  first 
case;  and  therefore  D  is  less  than  F. 
Therefore,  if  there  be  three,  &c.  Q.  E.  D. 


PROP.  XXI.  THEOR. 

If  there  be  three  magnitudes,  and  other  three,  which 
have  the  same  ratio  taken  two  and  two,  but  in  a  cross 
order,  if  the  first  magnitude  be  greater  than  the  third, 
the  fourth  shall  be  greater  than  the  sixth ;  and  if  equal, 
equal;  and  if  less,  less.* 

Let  A,  B,  C  be  three  magnitudes,  and  D,  E,  F  other  three, 
which  have  the  same  ratio,  taken  two  and  two,  but  in  a  cross 
order,  viz.  as  A  is  to  B,  so  is  E  to  F,  and  as  B  is 
to  C,  so  is  D  to  E.  If  A  be  greater  than  C,  D 
shall  be  greater  than  F;  and  if  equal,  equal;  and 
if  less,  less. 

Because  A  is  greater  than  C,  and  B  is  any  other 
magnitude,  A  has  to  B  a  greater  ratio  (8.  5.)  than 
C  has  to  B:  but  as  E  to  F,  so  is  A  to  B;  therefore 
(13.  5.)  E  has  to  F  a  greater  ratio  than  C  to  B:  and 
because  B  is  to  C,  as  D  to  E,  by  inversion,  C  is  to 


A     B 


*  See  Note, 


118 


THE  ELEMENTS  OF  EUCLID. 


BOOK  V. 


B,  as  E  to  D:  and  E  was  shown  to  have  to  F  a     D     E     F 

greater  ratio  than  C  to  B;  therefore  E  has  to  F  a 

greater  ratio  than  E  to  D  (Cor.  13.  5.);  but  the 

magnitude  to  which  the  same  has  a  greater  ratio 

than  it  has  to  another,  is  the  lesser  of  the  two 

(10.  5.);  F  therefore  is  less  than  D,  tha.  is,  D  is 

greater  than  F. 

Secondly,  Let  A  be  equal  to  C,  D  shall  be  equal  to  F.  Because 
A  and  C  are  equal,  A  is  (7.  5.)  to  B,  as  C  is  to  B:  but  A  is  to  B, 
as  E  to  Fj  and  C  is  to  B,  as  E  to  D; 
wherefore  E  is  to  F,  as  E  to  D  (11. 
5.);  and  therefore  D  is  equal  to  F 
(9.  5.). 

Next,  Let  A  be  less  than  C^  D 
shall  be  less  than  F;  for  C  is  greater 
than  A,  and,  as  was  shown,  C  is  to 
B,  as  E  to  D;  and  in  like  manner  B 
is  to  A,  as  F  to  E^  therefore  F  is 
greater  than  D,  by  case  first;  and 
therefore  D  is  less  than  F.  There- 
fore, if  there  be  three,  Sec.  Q.  E.  D. 


A 
D 


B 
E 


C 

F 


A 
D 


E 


C 
F 


PROP.  XXn.  THEOR. 

If  there  be  any  number  of  magnitudes,  and  as  many 
others,  which,  taken  two  and  two  in  order,  have  the 
same  ratio;  the  first  shall  have  to  the  last  of  the  first 
magnitudes  the  same  ratio  which  the  first  of  the  others 
has  to  the  last.  N.  B.  This  is  usually  cited  by  the  words 
^'ex  cBquali,'^^  or  "ex  ceguo*^^^ 

First,  Let  there  be  three  magnitudes  A,  B,  C,  and  as  many 
others  D,  E,  F,  which  taken  two  and  two,  have  the  same  ratio, 
that  is,  such  that  A  is  to  B,  as  D  to  E;  and  as  B  is  to  C,  so  is  E 
to  F;  A  shall  be  to  C,  as  D  to  F. 

Take  of  A  and  D  any  equimultiples  whatever  G  and  H;  and  of 
B  and  E  any  equimultiples  what- 
ever K  and  L;  and  of  C  and  F  any 
whatever  M  and  N:  then,  because 
A  is  to  B,  as  D  to  E,  and  that  G, 
H  are  equimultiples  of  A,  D,  and 
K,  L  equimultiples  of  B,  E;  as  G 

is  to  K,  so  is  (4.  5.)  H  to  L.     For     A      B      C      D      E      F 
the  same  reason,  K  is  to  M,  as  L 


*  See  Note. 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


119 


to  N:  and  because  there  are  three  G  K  M  H  L  N 
magnitudes  G,  K,  M,  and  other 
three  H,  L,  N,  which,  two  and 
two,  have  the  same  ratio:  if  G  be 
greater  than  M,  H  is  greater  than 
N|  and  if  equal,  equalj  and  if  less, 
less  (20.  5.);  and  G,H  are  any  equi- 
multiples whatever  of  A,  D,  and 
M,  N  are  any  equimultiples  what- 
ever of  C,  F.  Therefore,  (5.  def. 
5.),  as  A  is  to  C,  so  is  D  to  F. 

Next,  Let  there  be  four  magnitudes.  A,  B,  C,  D,  and  other  four, 
E,  F,  G,  H,  which,  two  and  two,  have  the  same 
ratio,  viz.  as  A  is  to  B,  so  is  E  to  F,  and  as  B  to 
C,  so  F  to  G;  and  as  C  to  D,  so  G  to  H:  A  shall 
be  to  D,  as  E  to  H. 

Because  A,  B,  C  are  three  magnitudes,  and  E,  F,  G  other  three, 
which,  taken  two  and  two,  have  the  same  ratio;  by  the  foregoing 
case,  A  is  to  C,  as  E  to  G.  But  C  is  to  D,  as  G  is  to  H;  where- 
fore again,  by  the  first  case,  A  is  to  D,  as  E  to  H:  and  so  on, 
whatever  be  the  number  of  magnitudes.  Therefore,  if  there  be 
any  number,  &c.  Q.  E.  D. 


A.  B.  C.  D. 
E.  F.  G.  H. 


PROP.  XXIII.  THEOR. 

If  there  be  any  number  of  magnitudes,  and  as  many 
others,  which,  taken  two  and  two,  in  a  cross  order, 
have  the  same  ratio;  the  first  shall  have  to  the  last  of 
the  first  magnitudes  the  same  ratio  which  the  first  of 
the  others  has  to  the  last.  N.  B.  This  is  usually  cited 
by  the  ivords  "'ex  cequali  in  proportione  perturhata;^'^  or^ 
"  ex  CBqiio  perturhatoy^ 

First,  Let  there  be  three  magnitudes  A,  B,  C,  and  other  three 
D,  E,  F,  which,  taken  two  and  two  in  a  cross  order,  have  the  same 
ratio,  that  is,  such  that  A  is  to  B,  as  E  to  F;  and  as  B  is  to  C,  so 
is  D  to  E:  A  is  to  C,  as  D  to  F. 

Take  of  A,  B,  D  any  equimultiples  whatever  G,  H,  K;  and  of 
C,  E,  F  any  equimultiples  whatever  L,  M,  N;  and  because  G,  H 
are  equimultiples  of  A,  B,  and 
that  magnitudes  have  the  same 
ratio  which  their  equimultiples 
have  (15.  5.);  as  A  is  to  B,  so  is 
G  to  H.  And  for  the  same  rea- 
son, as  E  is  to  F,  so  is  M  to  N; 
but  as  A  is  to  B,  so  is  E  to  F;  as 
therefore  G  is  to  H,  so  is  M  to  N 
(11.  5.).     And  because  as  B  is  to 


ABC 


DEF 


*  See  Note. 


120 


THE  ELEMENTS  OF  EUCLID. 


BOOK  V. 


C,  so  is  D  to  E,  and  that  H,  KGHL  KMN 

are  equimultiples  of  B,  D,  and  L, 
M,  of  C,  E|  as  H  is  to  L,  so  is 
(4,  5.)  K  to  M:  and  it  has  been 
shown,  that  G  is  to  H,  as  M  to 
N;  then,  because  there  are  three 
magnitudes  G,  H,  L,  and  other 
three  K,  M,  N,  which  have  the 
same  ratio  taken  two  and  two  in 
a  cross  order;  if  G  be  greater 
than  L,  K  is  greater  than  N;  and 

if  equal,  equal;  and  if  less,  less  (21.  5.);  and  G,  K  are  any  equi- 
multiples whatever  of  A,  D;  and  L,  N  any  whatever  of  C,  F;  as, 
therefore,  A  is  to  C,  so  is  D  to  F. 

Next,  Let  there  be  four  magnitudes.  A,  B,  C,  D,  and  other  four 
E,  F,  G,  H,  which,  taken  two  and  two  in  a  cross 
order,  have  the  same  ratio,  viz.  A  to  B,  as  G  to  H; 
B  to  C,  as  F  to  G;  and  C  to  D,  as  E  to  F:  A  is  to 


A.  B.  C.  D. 
E.  F.  G.  H. 


D,  as  E  to  H. 

Because  A,  B,  C  are  three  magnitudes,  and  F,  G,  H  other  three, 
which,  taken  two  and  two  in  a  cross  order,  have  the  same  ratio; 
by  the  first  case,  A  is  to  C,  as  F  to  H:  but  C  is  to  D,  as  E  is  to 
F;  wherefore  again,  by  the  first  case,  A  is  to  D,  as  E  to  H:  and 
so  on,  whatever  be  the  number  of  magnitudes.  Therefore,  if  there 
be  any  number,  &c.  Q.  E.  D. 

PROP.  XXIV.  THEOR. 

If  the  first  has  to  the  second  the  same  ratio  which 
the  third  has  to  the  fourth;  and  the  fifth  to  the  second, 
the  same  ratio  which  the  sixth  has  to  the  fourth;  the 
first  and  fifth  together  shall  have  to  the  second,  the 
same  ratio  which  the  third  and  sixth  together  have  to 
the  fourth.* 

Let  AB  the  first,  have  to  C  the  second,  the  same  ratio  which 
DE  the  third,  has  to  F  the  fourth;  and  let  BG  the  fifth,  have  to 
C  the  second,  the  same  ratio  which  EH 
the  sixth,  has  to  F  the  fourth:  AG,  the  G 
first  and  fifth  together,  shall  have  to  C 
the  second,  the  same  ratio  which  DH, 
the  third  and  sixth  together,  has  to  F  -^ 
the  fourth. 

Because  BG  is  to  C,  as  EH  to  F;  by 
inversion,  C  is  to  BG,  as  F  to  EH:  and 
because  as  AB  is  to  C,  so  is  DE  to  F; 
and  as  C  to  BG,  so  F  to  EH;  ex  sequali 
(22.  5.),  AB  is  to  BG,  as  DE  to  EH: 
and  because  these  magnitudes  are  pro- 
portionals, they  shall  likewise  be  pro-  ^  C 
portionals  when  taken  jointly  (18.  5.): 

«  See  Note. 


H 


E- 


D      F 


.iU; 


BOOK  V. 


THE  ELEMENTS  OF  EUCLID. 


121 


as,  therefore,  AG  is  to  GB,  so  is  DH  to  HE;  but  as  GB  to  C,  so 
is  HE  to  F.  Therefore,  ex  dequali  (22.  5.)?  as  AG  is  to  C,  so  is 
DH  to  F.  Wherefore,  if  the  first.  Sec.  Q.  E.  D. 
"  CoR.  1.  If  the  same  hypothesis  be  made  as  in  the  proposition, 
the  excess  of  the  first  and  fifth  shall  be  to  the  second,  as  the  excess 
of  the  third  and  sixth  to  the  fourth.  The  demonstration  of  this 
is  the  same  with  that  of  the  proposition,  if  division  be  used  in- 
stead of  composition. 

Cor.  2.  The  proposition  holds  true  of  two  ranks  of  magni- 
tudes, whatever  be  their  number,  of  which  each  of  the  first  rank 
has  to  the  second  magnitude  the  same  ratio  that  the  correspond- 
ing one  of  the  second  rank  has  to  a  fourth  magnitude;  as  is  ma- 
nifest. 

PROP.  XXV.  THEOR. 

If  four  magnitudes  of  the  same  kind  are  proportion- 
als, the  greatest  and  least  of  them  together  are  greater 
than  the  other  two  together. 

Let  the  four  magnitudes  AB,  CD,  E,  F,  be  proportionals;  viz. 
AB  to  CD,  as  E  to  F;  and  let  AB  be  the  greatest  of  them,  and 
consequently  F  the  least  (A.  &  14.  15.).  AB,  together  with  F, 
are  greater  than  CD,  together  with  E. 

Take  AG  equal  to  E,  and  CH  equal  to  F:  then,  because  as  AB 
is  to  CD,  so  is  E  to  F,  and  that  AG  is  equal  to  E,  and  CH  equal 
to  F;  AB  is  to  CD,  as  AG  to  CH.  And  3 
because  AB  the  whole  is  to  the  whole 
CD,  as  AG  is  to  CH,  likewise  the  re- 
mainder GB  shall  be  to  the  remainder  G-- 
HD,  as  the  whole  AB  is  to  the  whole 
(19.  5.)  CD:  but  AB  is  greater  than  CD, 
therefore  (A.  5.)  GB  is  greater  than  HD:  H 

and  because  AG  is  equal  to  E,  and  CH 
to  F;  AG  and  F  together  are  equal  to 
CH  and  E  together.  If,  therefore,  to 
the  unequal  magnitudes  GB,  HD,  of 
which  GB  is  the  greater,  there  be  added  equal  magnitudes,  viz. 
to  GB  the  two  AG  and  F,  and  CH  and  E  to  HD;  AB  and  F  to- 
gether are  greater  than  CD  and  E.  Therefore,  if  four  magni- 
tudes, 8cc.     Q.  E.  D. 

PROP.  F.  THEOR. 

Ratios  which  are  compounded  of  the  same  ratios, 
are  the  same  with  one  another.* 

Let  A  be  to  B,  as  D  to  E;  and  B  to  C,  as  E  to  F:  the  ratio 
which  is  compounded  of  the  ratios  of  A 
to  B,  and  B  to  C,  which  by  the  definition 
of  compound  ratio,  is  the  ratio  of  A  to  C, 
is  the  same  with  the  ratio  of  D  to  F, 
which  by  the  same  definition  is  com- 
pounded of  the  ratios  of  D  to  E,  and  E, 
to  F. 

'^  See  Note. 

Q 


D 
I 


E     F 


A.  B.  C. 

D.  E.  F. 


122  THE  ELEMENTS  OF  EUCLID.  BOOK  V. 

Because  there  are  three  magnitudes,  A,  B,  C,  and  three  others 
D,  E,  F,  which,  taken  two  and  two  in  order,  have  the  same  ratio: 
ex  sequali^  A  is  to  C,  as  D  to  F  (22.  5.). 

Next,  Let  A  be  to  B,  as  E  to  F,  and  B  to  C,  as  D  to  Ej  there- 
fore, ex  dequali  in  proportions  perlurbata 
(23.  5.),  A  is  to  C,  as  D  to  F;  that  is,  the 
ratio  of  A  to  C,  which  is  compounded  of 
the  ratios  of  A  to  B,  and  B  to  C,  is  the 
same  with  the  ratio  of  D  to  F,  which  is 
compounded  of  the  ratios  of  D  to  E,  and 
E  to  F;  and  in  like  manner  the  propositions  may  be  demonstrated, 
v/hatever  be  the  number  of  ratios  in  either  case. 

PROP.  G.  THEOPv. 

If  several  ratios  be  the  same  with  several  ratios,  each 
to  each:  the  ratio  which  is  compounded  of  ratios  which 
are  the  same  with  the  first  ratios,  each  to  each;  is  the 
same  with  the  ratio  compounded  of  ratios  which  are 
the  same  with  the  other  ratios,  each  to  each.^ 

Let  A  be  to  B,  as  E  to  F;  and  C  to  D,  as  G  to  H:  and  let  A  be 
to  B,  as  K  to  L;  and  C  to  D,  as  L  to  M:  then  the  ratio  of  K  to  M, 
by  the  definition  of  compound  ratio, 
is  compounded  of  the  ratios  of  K  to 
L,  and  L  to  M,  which  are  the  same 
with  the  ratios  of  A  to  B,  and  C  to 
D;  and  as  E  to  F,  so  let  N  be  to  O; 
and  as  G  to  H,  so  let  O  be  to  P;  then  the  ratio  of  N  to  P  is  com- 
pounded of  the  ratios  of  N  to  O,  and  O  to  P,  which  are  the  same 
with  the  ratios  of  E  to  F,  and  G  to  H:  and  it  is  to  be  shown  that 
the  ratio  of  K  to  M,  is  the  same  with  the  ratio  of  N  to  P,  or 
that  K  is  to  M,  as  N  to  P. 

Because  K  is  to  L,  as  (A  to  B,  that  is,  as  E  to  F,  that  is,  as) 
N  to  O;  and  as  L  to  M,  so  is  (C  to  D,  and  so  is  G  to  H,  and  so 
is)  O  to  P:  ex  dequali  (22.  5.)  K  is  to  M,  as  N  to  P.  Therefore, 
if  several  ratios,  Sec.  Q.  E.  D. 

PROP.  H.  THEOR. 

If  a  ratio  compounded  of  several  ratios  be  the  same 
with  a  ratio  compounded  of  any  other  ratios,  and  if  one 
of  the  first  ratios,  or  a  ratio  compounded  of  any  of  the 
first,  be  the  same  with  one  of  the  last  ratios,  or  with 
the  ratio  compounded  of  any  of  the  last ;  then  the  ratio 
compounded  of  the  remaining  ratios  of  the  first,  or  the 
remaining  ratio  of  the  first,  if  but  one  remain,  is  the 
same  with  the  ratio  compounded  of  those  remaining  of 
the  last,  or  with  the  remaining  ratio  of  the  last.* 

*  See  Note, 


A.  B.  C.  D.  K.  L.  M. 
E.  F.  G.  H.  N.  O.  P. 


.7/' 


BOOK  V.  THE  ELEMENTS  OF  EUCLID.  123 

Let  the  first  ratios  be  those  of  A  to  B,  B  to  C,  C  to  D,  D  to  E, 
and  E  to  F:  and  let  the  other  ratios  be  those  of  G  to  H,  H  to  K, 
K  to  L,  and  L  to  M:  also,  let  the  ratio  of  A  to  F,  which  is  com- 
pounded of*  the  first  ratios,  be  the  same 


A.  B.C.D.E.F. 
G.H.K.L.  M. 


with  the  ratio  of  G  to  M,  which  is  com- 
pounded of  the  other  ratiosj  and  be- 
sides, let  the  ratio  of  A  to  D,  which  is 
compounded  of  the  ratios  of  A  to  B,  B 
to  C,  C  to  D,  be  the  same  with  the  ratio  of  G  to  K,  which  is  com- 
pounded of  the  ratios  of  G  to  H,  and  H  to  K,  then  the  ratio  com- 
pounded of  the  remaining  first  ratios,  to  wit,  of  the  ratios  of  D  to 
E,  and  E  to  F,  which  compounded  ratio  is  the  ratio  of  D  to  F,  is 
the  same  with  the  ratio  of  K  to  M,  which  is  compounded  of  the 
remaining  ratios  of  K  to  L,  and  L  to  M  of  the  other  ratios. 

Because,  by  the  hypothesis,  A  is  to  D,  as  G  to  K,  by  inversion 
(B.  5.),  D  is  to  A,  as  K  to  G;  and  as  A  is  to  F,  so  is  G  to  M; 
therefore  (22.  5.),  ex  eequali^  D  is  to  F,  as  K  to  M.  If  therefore  a 
ratio  which  is,  &c.  Q.  E.  D. 

PROP.  K.  THEOR. 

If  there  be  any  number  of  ratios,  and  any  number  of 
other  ratios  such,  that  the  ratio  compounded  of  ratios 
which  are  the  same  with  the  first  ratios,  each  to  each, 
is  the  same  with  the  ratio  compounded  of  ratios  which 
are  the  same,  each  to  each,  with  the  last  ratios;  and 
if  one  of  the  first  ratios,  or  the  ratio  which  is  com- 
pounded of  ratios  which  are  the  same  with  several  of 
the  first  ratios,  each  to  each,  be  the  same  with  one  of 
the  last  ratios,  or  with  the  ratio  compounded  of  ratios 
which  are  the  same,  each  to  each,  with  several  of  the 
last  ratios :  then  the  ratio  compounded  of  ratios  which 
are  the  same  with  the  remaining  ratios  of  the  first, 
each  to  each,  or  the  remaining  ratio  of  the  first,  if  but 
one  remain;  is  the  same  with  the  ratio  compounded  of 
ratios  which  are  the  same  with  those  remaining  of  the 
last,  each  to  each,  or  with  the  remaining  ratio  of  the 
last.t 

Let  the  ratios  of  A  to  B,  C  to  D,  E  to  F,  be  the  first  ratios; 
and  the  ratios  of  G  to  H,  K  to  L,  M  to  N,  O  to  P,  Q  to  R,  be  the 
other  ratios:  and  let  A  be  to  B,  as  S  to  T;  and  C  to  D,  as  T  to 
V;  and  E  to  F,  as  V  to  X:  therefore,  by  the  definition  of  com- 
pound ratio,  the  ratio  of  S  to  X  is  compounded  of  the  ratios  of  S 

Definition  of  compound  ratio.  t  See  Note. 


124  THE  ELEMENTS  OF  EUCLID.  BOOK  V. 


h, 

k, 

1, 

A 

.B; 

c, 

D; 

E, 

F. 

s, 

T, 

V, 

X. 

G, 

H: 

e,  f, 

K, 

U 

M, 

m, 

P;  Q, 

o,  p. 

R 

Y 

>z, 

a, 

b,  c, 

d. 

to  T,  T  to  V,  and  V  to  X,  which  are  the  same  with  the  ratios  of 
A  to  B,  C  to  D,  E  to  F,  each  to  each;  also,  as  G  to  H,  so  let  Y  be 
to  Z;  and  K  to  L,  as  Z  to  a;  M  to  N,  as  a  to  b,  O  to  P,  as  b  to  c; 
and  Q  to  R,  as  c  to  d:  therefore,  by  the  same  definition,  the  ratio 
of  Y  to  d  is  compounded  of  the  ratios  of  Y  to  Z,  Z  to  a,  a  to  b,  b 
to  c,  and  c  to  d,  which  are  the  same,  eacli  to  each,  with  the  ratios 
of  G  to  H,  K  to  L,  M  to  N,  O  to  P,  and  Q  to  R:  therefore,  by  the 
hypothesis,  S  is  to  X,  as  Y  to  d;  also,  let  the  ratio  of  A  to  B,  that 
is  the  ratio  of  S  to  T,  which  is  one  of  the  first  ratios,  be  the  same 
with  the  ratio  of  e  to  g",  which  is  compounded  of  the  ratios  of  e  to 
f,  and  f  to  g,  which,  by  the  hypothesis,  are  the  same  with  the  ratios 
of  G  to  H,  and  K  to  L,  two  of  the  other  ratios;  and  let  the  ratio 
of  h  to  1  be  that  which  is  compounded  of  the  ratios  of  h  to  k,  and 
k  to  1,  which  are  the  same  with  the  remaining  first  ratios,  viz.  of 
C  to  D,  and  E  to  F;  also,  let  the  ratio  of  m  to  p,  be  that  which 
is  compounded  of  the  ratios  of  m  to  n,  n  to  o,  and  o  to  p,  which 
are  the  same,  each  to  each,  with  the  remaining  other  ratios,  viz. 
of  M  to  N,  O  to  P,  and  Q  to  R:  then  the  ratio  of  h  to  1  is  the  same 
with  the  ratio  of  m  to  p,  or  h  is  to  1,  as  m  to  p. 


h, 

k, 

1, 

A 

,B; 

c, 

D: 

E, 

F. 

s, 

T, 

V, 

X. 

G, 

H; 

L; 

M, 

N; 
m, 

o, 

P5  Q, 

o,  p. 

R 

.  Y 

>Z 

.a, 

b,  c, 

d. 

Because  e  is  to  f,  as  (G  to  H,  that  is,  as)  Y  to  Z;  and  f  is  to  g, 
as  (K  to  L,  that  is,  as)  Z  to  a;  therefore,  ex  aequali^  e  is  to  g,  as 
Y  to  a:  and  by  the  hypothesis,  A  is  to  B,  that  is,  S  to  T,  as  e  to 
g;  wherefore  S  is  to  T,  as  Y  to  a;  and  by  inversion,  T  is  to  S,  as 
a  to  Y;  and  S  is  to  X,  as  Y  to  d:  therefore,  ex  sequali,  T  is  to  X, 
as  a  to  d:  also,  because  h  is  to  k,  as  (C  to  D,  that  is,  as)  T  to  V: 
and  k  is  to  1,  as  (E  to  F,  that  is,  as)  V  to  X;  therefore,  ex  sequali^ 
h  is  to  1,  as  T  to  X:  in  like  manner,  it  may  be  demonstrated,  that 
m  is  to  p,  as  a  to  d:  and  it  has  been  shown,  that  T  is  to  X,  as  a 
to  d;  therefore  (l  1.  5.)  h  is  to  1,  as  m  to  p.  Q.  E.  D. 

The  propositions  G  and  K  are  usually,  for  the  sake  of  brevity, 
expressed  in  the  same  terms  with  propositions  F  and  H:  and 
therefore  it  was  proper  to  show  the  true  meaning  of  them  when 
they  are  so  expressed;  especially  since  they  are  very  frequently 
made  use  of  by  geometers. 


THE 


ELEMENTS  OF  EUCLID* 


BOOK  yr. 


DEFIISriTIONS. 


I. 

Similar  rectilineal  figures  are  those 
which  have  their  several  angles 
equal,  each  to  each,  and  the  sides 
about  the  equal  angles  propor- 
tionals. 

II. 

"  Reciprocal  figures,  viz.  triangles  and  parallelograms  are  such 
"  as  have  their  sides  about  two  of  their  angles  proportionals 
"  in  such  manner,  that  a  side  of  the  first  figure  is  to  a  side  of 
"  the  other,  as  the  remaining  side  of  this  other  is  to  the  re- 
"  maining  side  of  the  first."* 

III. 

A  straight  line  is  said  to  be  cut  in  extreme  and  mean  ratio,  when 
the  whole  is  to  the  greater  segment,  as  the  greater  segment  is 
to  the  less. 

IV. 

The  altitude  of  any  figure  is  the  straight  line 
drawn  from  its  vertex  perpendicular  to  the 
base. 


PROP.  I.  THEOR. 

Triangles  and  parallelograms  of  the  same  altitude 
are  one  to  another  as  their  bases.* 

Let  the  triangles  ABC,  ACD,  and  the  parallelograms  EC,  CF, 
have  the  same  altitude,  viz.  the  perpendicular  drawn  from  the 
point  A  to  BD  :  then,  as  the  base  BC  is  to  the  base  CD,  so  is  the 
triangle  ABC  to  the  triangle  ACD,  and  the  parallelogram  EC  to 
the  parallelogram  CF. 

*  See  Note. 


126  THE  ELEMENTS  OF  EUCLID.  BOOK  VI. 

Produce  BD  both  ways  to  the  points  H,  L,  and  take  any  num- 
ber of  straight  lines  BG,  GH,  each  equal  tQ  the  base  BC;  and 
DK,  KL,  any  number  of  them,  each  equal  to  the  base  CD,  and 
join  AG,  AH,  AK,  AL:  then,  because  CB,  BG,  GH  are  all  equal, 
the  triangles  AHG,  AGB,  ABC  are  all  equal  (38.  1.):  therefore, 
whatever  multiple  the  base  IJC  is  of  the  base  BC,  the  same  mul- 
tiple is  the  triangle  AHC  of  the  triangle  ABC|  for  the  same  rea- 
son, whatever  multiple  the  E  A  F 
base  LC  is  of  the  base  CD, 
the  same  multiple  is  the  tri- 
angle ALC  of  the  triangle 
ADC:  and  if  the  base  HC 
be  equal  to  the  base  CL,  the 
triangle  AHC  is  also  equal 
to  the  triangle  ALC  (38. 1 .); 
and  if  the  base  HC  be  H  G  B  C  D  K  L 
greater  than  the  base  CL,  likewise  the  triangle  AHC  is  greater 
than  the  triangle  ALC^  and  if  less,  less:  therefore,  since  there  are 
four  magnitudes,  viz.  the  two  bases  BC,  CD,  and  the  two  triangles 
ABC,  ACD,  and  of  the  base  BC  and  the  triangle  ABC  the  first 
and  third,  any  equimultiples  whatever  have  been  taken,  viz.  the 
base  HC  and  triangle  AHC^  and  of  the  base  CD  and  triangle 
ACD,  the  second  and  fourth,  have  been  taken  any  equimultiples 
whatever,  viz.  the  base  CL,  and  triangle  ALC^  and  that  it  has 
been  shown,  that,  if  the  base  HC  be  greater  than  the  base  CL,  the 
triangle  AHC  is  greater  than  the  triangle  ALC^  and  if  equal, 
equal;  and  if  less,  less|  therefore  (5.  def.  5.)  as  the  base  BC  is  to 
the  base  CD,  so  is  the  triangle  ABC  to  the  triangle  ACD. 

And  because  the  parallelogram  CE  is  double  of  the  triangle 
ABC  (41. 1 .)  and  the  parallelogram  CF  double  of  the  triangle  ACD, 
and  that  magnitudes  have  the  same  ratio  which  their  equimulti- 
ples have  (15.  5.):  as  the  triangle  ABC  is  to  the  triangle  ACD,  so 
is  the  parallelogram  EC  to  the  parallelogram  CF;  and  because  it 
has  been  shown,  that,  as  the  base  BC  is  to  the  base  CD,  so  is  the 
triangle  ABC  to  the  triangle  ACD;  and  as  the  triangle  ABC  to 
the  triangle  ACD,  so  is  the  parallelogram  EC  to  the  parallelo- 
gram CF;  therefore  as  the  base  BC  is  to  the  base  CD,  so  is  (l  I. 
5.)  the  parallelogram  EC  to  the  parallelogram  CF.  Wherefore, 
triangles,  8cc.  Q.  E.  D. 

Cor.  From  this  it  is  plain,  that  triangles  and  parallelograms 
that  have  equal  altitudes,  are  one  to  another  as  their  bases. 

Let  the  figures  be  placed  so  as  to  have  their  bases  in  the  same 
straight  line;  and  having  drawn  perpendiculars  from  the  vertices 
of  the  triangles  to  the  bases,  the  straight  line  which  joins  the  ver- 
tices is  parallel  to  that  in  which  their  bases  are  (33.  1.),  because 
the  perpendiculars  are  both  equal  and  parallel  to  one  another: 
then,  if  the  same  construction  be  made  as  in  the  proposition,  the 
demonstration  will  be  the  same. 


BOOK  VI. 


THE  ELEMENTS  OF  E,UCLID. 


127 


PROP.  II.  THEOR. 

If  a  straight  line  be  drawn  parallel  to  one  of  the 
sides  of  a  triangle,  it  shall  cut  the  other  sides,  or  those 
produced,  proportionally;  and  if  the  sides,  or  the  sides 
produced,  be  cut  proportionally,  the  straight  hue  which 
joins  the  points  of  section  shall  be  parallel  to  the  re- 
maining side  of  the  triangle.* 

Let  DE  be  drawn  parallel  to  BC,  one  of  the  sides  of  the  trian- 
gle ABC;  BD  is  to  DA,  as  CE  to  EA. 

Join  BE,  CD;  then  the  triangle  BDE  is  equal  to  the  triangle 
CDE  (37.  I.),  because  they  are  on  the  same  base  DE,  and  between 
the  same  parallels  DE,  BC;  ADE  is  another  triangle,  and  equal 
magnitudes  have  to  the  same  the  same  ratio  (7.  5.);  therefore,  as 
the  triangle  BDE  to  the  triangle  ADE,  so  is  the  triangle  CDE  to 
the  triangle  ADE;  but  as  the  triangle  BDE  to  the  triangle  ADE, 
so  is  (l.  6.)  BD  to  DA,  because  having  the  same  altitude,  viz.  the 
perpendicular  drawn  from  the  point  E  to  AB,  they  are  to  one 
another  as  their  bases;  and  for  the  same  reason,  as  the  triangle 
CDE  to  the  triangle  ADE,  so  is  CE  to  EA.  Therefore,  as  BD 
to  DA,  so  is  CE  to  EA  (11.  5.). 

Next,  let  the  sides  AB,  AC  of  the  triangle  ABC,  or  these  pro- 


D 


B  C 

duced,  be  cut  proportionally  in  the  points  D,  E,  that  is,  so  that 
BD  be  to  DA,  as  CE  to  EA,  and  join  DE;  DE  is  parallel  to  BC. 
The  same  construction  being  made;  because  as  BD  to  DA,  so 
is  CE  to  EA;  and  as  BD  to  DA,  so  is  the  triangle  BDE  to  the 
triangle  ADE  (l.  6.);  and  as  CE  to  EA,  so  is  the  triangle  CDE 
to  the  triangle  ADE;  therefore  the  triangle  BDE  is  to  the  trian- 
gle ADE,  as  the  triangle  CDE  to  the  triangle  ADE;  that  is,  the 
triangles  BDE,  CDE  have  the  same  ratio  to  the  triangle  ADE; 
and  therefore  (9.  5.)  the  triangle  BDE  is  equal  to  the  triangle 
CDE;  and  they  are  on  the  same  base  DE;  but  equal  triangles  on 
the  same  base  are  between  the  same  parallels  (39.  1.);  therefore 
DE  is  parallel  to  BC.    Wherefore,  if  a  straight  line,  Sec.  Q.  E.  D. 

*  See  Note. 


128  THE  ELEMENTS  OF  EUCLID.  BOOK  VI. 


PROP.  III.  THEOR. 

If  the  angle  of  a  triangle  be  divided  into  two  equal 
angles,  by  a  straight  line  which  also  cuts  the  base;  the 
segments  of  the  base  shall  have  the  same  ratio  which 
the  other  sides  of  the  triangle  have  to  one  another: 
and  if  the  segments  of  the  base  have  the  same  ratio 
which  the  other  sides  of  the  triangle  have  to  one  ano- 
ther, the  straight  line  drawn  from  the  vertex  to  the 
point  of  section,  divides  the  vertical  angle  into  two 
equal  angles.^ 

Let  the  angle  BAC  of  any  triangle  ABC  be  divided  into  two 
equal  angles  by  the  straight  line  AD:  BD  is  to  DC,  as  BA  to 
AC. 

Through  the  point  C  draw  CE  parallel  (31.  1.)  to  DA,  and  let 
BA  produced  meet  CE  in  E.  Because  the  straight  line  AC  meets 
the  parallels  AD,  EC,  the  angle  ACE  is  equal  to  the  alternate 
angle  CAD  (29.  1.);  but  CAD,  by  the  hypothesis  is  equal  to  the 
angle  BAD;  wherefore  BAD  is  equal  to  the  angle  ACE.  Again, 
because    the    straight   line    BAE  E 

meets  the  parallels  AD,  EC,  the  .'^ 

outward  angle  BAD  is  equal  to 
the  inward  and  opposite  angle 
AEC:  but  the  angle  ACE  has 
been  proved  equal  to  the  angle 
BAD;  therefore  also  ACE  is 
equal  to  the  angle  AEC,  and  con- 
sequently the  side  AE  is  equal  to 
the  side  (6.  1.)  AC;  and  because 
AD  is  drawn  parallel  to  one  of  the  sides  of  the  triangle  BCE,  viz. 
to  EC,  BD  is  to  DC,  as  BA  to  AE  (2.  6.):  but  AE  is  equal  to  AC; 
therefore,  as  BD  to  DC,  so  is  BA  to  AC  (7.  5.). 

Let  now  BD  be  to  DC,  as  BA  to  AC,  and  join  AD;  the  angle 
BAC  is  divided  into  two  equal  angles  by  the  straight  line  AD. 

The  same  construction  being  made;  because,  as  BD  to  DC,  so 
is  BA  to  AC:  and  as  BD  to  I3C,  so  is  BA  to  AE  (2.  6.),  because 
AD  is  parallel  to  EC;  therefore  BA  is  to  AC,  as  BA  to  AE  (II. 
5.):  consequently  AC  is  equal  to  AE  (9.  5.),  and  the  angle  AEC 
is  therefore  equal  to  the  angle  ACE  (5.  1.):  but  the  angle  AEC 
is  equal  to  the  outward  and  opposite  angle  BAD:  and  the  angle 
ACE  is  equal  to  the  alternate  angle  CAD  (29.  1.):  wherefore  also 
the  angle  BAD  is  equal  to  the  angle  CAD:  therefore  the  angle 
BAC  is  cut  into  two  equal  angles  by  the  straight  line  AD.  There- 
fore, if  the  angle,  &c.   Q.  E.  D. 

*  See  Note, 


BOOK  VI.  THE  ELEMENTS  OF  EUCLID.  129 

PROP.  A.    THEOR. 

If  the  outward  angle  of  a  triangle  made  by  producing 
one  of  its  sides,  be  divided  into  two  equal  angles,  by  a 
straight  line  which  also  cuts  the  base  produced ;  the 
segments  between  the  dividing  line  and  the  extremities 
of  the  base  have  the  same  ratio  which  the  other  sides 
of  the  triangle  have  to  one  another :  and  if  the  seg- 
ments of  the  base  produced  have  the  same  ratio  which 
the  other  sides  of  the  triangle  have,  the  straight  line 
drawn  from  the  vertex  to  the  point  of  section  divides 
the  outward  angle  of  the  triangle  into  two  equal  an- 
gles. 

Let  the  outward  angle  CAE  of  any  triangle  ABC  be  divided 
into  two  equal  angles  by  the  straight  line  AD  which  meets  the 
base  produced  in  D:  BD  is  to  DC,  as  BA  to  AC. 

Through  C  draw  CF  parallel  to  AD  (31.  1.):  and  because  the 
straight  line  AC  meets  the  parallels  AD,  FC,  the  angle  ACF  is 
equal  to  the  alternate  angle  CAD  (29.  1.):  but  CAD  is  equal  to 
the  angle  DAE  (Hyp.)j  therefore  also  DAE  is  equal  to  the  angle 
ACF.  Again,  because  the  straight  line  FAE  meets  the  parallels 
AD,   FC,    the    outward    angle  E 

DAE  is  equal  to  the  inward  and 

opposite    angle  CFA:    but  the  -**- 

angle  ACF  has  been  proved 
equal  to  the  angle  DAE  ;  there-  )r^ 

fore  also  the  angle  ACF  is  equal 
to  the  angle  CFA,  and  conse- 
quently the  side  AF  is  equal  to 
the   side   AC   (6.   1.):    and  be-    ^  ^  ^ 

cause  AD  is  parallel  to  FC,  a  side  of  the  triangle  BCF,  BD  is  to 
DC,  as  BA  to  AF  (2.  6.):  but  AF  is  equal  to  AC^  as  therefore 
BD  is  to  DC,  so  is  BA  to  AC. 

Let  now  BD  be  to  DC,  as  BA  to  AC,  and  join  AD^  the  angle 
CAD  is  equal  to  the  angle  DAE. 

The  same  construction  being  made,  because  BD  is  to  DC,  as 
BA  to  AC;  and  that  BD  is  also  to  DC,  as  BA  to  AF  (11.  5.): 
therefore  I3A  is  to  AC,  as  BA  to  AF  (9.  5.);  wherefore  AC  is 
equal  to  AF  (5.  1.),  and  the  angle  AFC  equal  (o.  1.)  to  the  angle 
ACF;  but  the  angle  AFC  is  equal  to  the  outward  angle  EAD, 
and  the  angle  ACF  to  the  alternate  angle  CAD:  therefore  also 
EAD  is  equal  to  the  angle  CAD.  Wherefore,  if  the  outward,  Sec. 
Q.  E.  D. 

PROP.  IV.  THEOR. 

The  sides  about  the  equal  angles  of  equiangular  tri- 
angles are  proportionals;  and  those  which  are  opposite 
R 


130  THE  ELEMENTS  OF  EUCLID.  BOOR  VI. 

to  the  equal  angles  are  homologous  sides,  that  is,  are 
the  antecedents  or  consequents  of  the  ratios. 

Let  ABC,  DCE  be  equiangular  triangles,  having  the  angle 
ABC  equal  to  the  angle  DCE,  and  the  angle  ACB  to  the  angle 
DEC,  and  consequently  (32.  1.)  the  angle  BAC  equal  to  the  an- 
gle CDE.  The  sides  about  the  equal  angles  of  the  triangles 
ABC,  DCE  are  proportionals:  and  those  are  the  homologous 
sides  which  are  opposite  to  the  equal  angles. 

Let  the  triangle  DCE  be  placed  so  that  its  side  CE  may  be 
contiguous  to  BC,  and  in  the  same  straight  line  with  it:  and  be- 
cause the  angles  ABC,  ACB  are  together  less  than  two  right 
angles  (1^-  l-)*  ^BC,  and  DEC,  F 
Avhich  is  equal  to  ACB,  are  also  less 
than  two  right  angles;  wherefore 
BA,  ED  produced  shall  meet  (12.  A 
Ax.  l.)j  let  them  be  produced  and 
meet  in  the  point  F;  and  because 
the  angle  ABC  is  equal  to  the  angle 

DCE,  BF  is  parallel  (28.  1.)  to  CD.  

Again,  because  the  angle  ACB  is  B  C  E 

equal  to  the  angle  DEC,  AC  is  parallel  to  FE  (28.  1.):  therefore 
FACD  is  a  parallelogram;  and  consequently  AF  is  equal  to  CD, 
and  AC  to  FD  (34.  1.):  and  because  AC  is  parallel  to  FE,  one  of 
the  sides  of  the  triangle  FBE,  BA  is  to  AF,  as  BC  to  CE  (2.  6.): 
but  AF  is  equal  to  CD;  therefore  (7.  5.),  as  BA  to  CD,  so  is  BC 
to  CE;  and  alternately,  as  AB  to  BC  so  is  DC  to  CE,  (17.  1.): 
again,  because  CD  is  parallel  to  BF,  as  BC  to  CE,  so  is  FD  to 
DE  (2.  6.):  but  FD  is  equal  to  AC:  therefore,  as  BC  to  CE,  so  is 
AC  to  DE:  and  alternately,  as  BC  to  CA,  so  CE  to  DE:  therefore, 
because  it  has  been  proved  that  AB  is  to  BC,  as  DC  to  CE,  and 
as  BC  to  CA,  so  CE  to  ED,  ex  sequali^  (22.  5.)  BA  is  to  AC,  as 
CD  to  DE.     Therefore  the  sides,  Sec.  Q.  E,  D. 

PROP.  V.  THEOR. 

If  the  sides  of  two  triangles,  about  each  of  their  an- 
gles, be  proportionals,  the  triangles  shall  be  equiangular, 
and  have  their  equal  angles  opposite  to  the  homologous 
sides. 

Let  the  triangles  ABC  DEF  have  their  sides  proportionals,  so 
that  AB  is  to  BC,  as  DE  to  EF;  and  BC  to  CA,  as  EF  to  FD; 
and  consequently,  ex  dequcdi,  BA  to  AC,  as  ED  to  DF;  the  trian- 
gle ABC  is  equiangular  to  the  triangle  DEF,  and  their  equal  an- 
gles are  opposite  to  the  homologous  sides,  viz.  the  angle  ABC 
equal  to  the  angle  DEF,  and  BCA  to  EFD,  and  also  BAC  to 
EDF. 

At  the  points  E,  F,  in  the  straight  line  EF,  make  (23.  1.)  the 
angle  FEG  equal  to  the  angle  ABC,  and  the  angle  EFG  equal  to 


BOOK  VI. 


THE  ELEMENTS  OF  EUCLID. 


131 


BCA;   wherefore   the   remain-  A  „        D 

ing  angle  BAG  is  equal  to  the 
remaining  angle  EGF  (32.  1.), 
and  the  triangle  ABC  is  there- 
fore equiangular  to  the  triangle 
GEF;  and  consequently  they 
have  their  sides  opposite  to  the 
equal  angles  proportionals  (4.  G 

6.).  Wherefore,  as  AB  to  BC,  so  is  OE  to  EF;  but  as  AB  to 
BC,  so  is  DE  to  EF;  therefore  as  DE  to  EF,  so  (U.  5.)  GE  to 
EF:  therefore  DE  and  GE  have  the  same  ratio  to  EF,  and  conse- 
quently are  equal  (9.  5.):  for  the  same  reason,  DF  is  equal  to  FG: 
and  because  in  the  triangles  DEF,  GEF,  DE  is  equal  to  EG,  and 
EF  common,  the  two  sides  DE,  EF  are  equal  to  the  two  GE,  EF, 
and  the  base  DF  is  equal  to  the  base  GF:  therefore  the  angle 
DEF  is  equal  (8.  1.)  to  the  angle  GEF,  and  the  other  angles  to 
the  other  angles  which  are  subtended  by  the  equal  sides  (4.  1.): 
wherefore  the  angle  DFE  is  equal  to  the  angle  GFE,  and  EDF  to 
EGF:  and  because  the  angle  DEF  is  equal  to  the  angle  GEF,  and 
GEF  to  the  angle  ABC;  therefore  the  angle  ABC  is  equal  to  the 
angle  DEF:  for  the  same  reason  the  angle  ACB  is  equal  to  the 
angle  DFE,  and  the  angle  at  A  to  the  angle  at  D.  Therefore  the 
triangle  ABC  is  equiangular  to  the  triangle  DEF.  Wherefore, 
if  the  sides,  &c.  Q.  E.  D. 


PROP.  VI.  THEOR. 

If  two  triangles  have  one  angle  of  the  one  equal  to 
one  angle  of  the  other,  and  the  sides  about  the  equal 
angles  proportionals,  the  triangles  shall  be  equiangular, 
and  shall  have  those  angles  equal  which  are  opposite  to 
the  homologous  sides. 

Let  the  triangles  ABC,  DEF  have  the  angle  BAC  in  the  one 
equal  to  the  angle  EDF  in  the  other,  and  the  sides  about  those 
angles  proportionals;  that  is,  BA  to  AC,  as  ED  to  DF;  the  tri- 
angles ABC,  DEF  are  equiangular,  and  have  the  angle  ABC 
equal  to  the  angle  DEF,  and  ACB  to  DFE. 

At  the  points  D,  F,  in  the  straight  line  DF,  make  (23.  1.)  the 
angle  FDG  equal  to  either  of  the  angles  BAC,  EDF;  and  the 
angle  DFG  equal  to  the  angle  A 
ACB;  wherefore  the  remaining 
angle  at  B  is  equal  to  the  re- 
maining one  at  G  (32.  I.),  and 
consequently  the  triangle  ABC 
is  equiangular  to  the  triangle 
DGF;  and  therefore  as  BA  to 
AC,  so  is  (4.  6.)  GD  to  DF;  but, 
by  the  hypothesis,  as  BA  to  AC,     B  C      E  F 

so  is  ED  to  DF;  as  therefore  ED  to  DF,  so  is  ( 1 1 ,  5.)  GD  to  DF; 


132  THE  ELEMENTS  OF  EUCLID.  BOOK  VI. 

wherefore  ED  is  equal  (9.  5.)  to  DG;  and  DF  is  common  to  the 
two  triangles  EDF,  GDF;  therefore  the  two  sides  ED,  DF  are 
equal  to  the  two  sides  GD,  DF:  and  the  angle  EDF  is  equal 
to  the  angle  GDF;  wherefore  the  base  EF  is  equal  to  the  base 
FG  (4.  1.),  and  the  triangle  EDF  to  the  triangle  GDF,  and  the  re- 
maining angles  to  the  remaining  angles,  each  to  each,  which  are 
subtended  by  the  equal  sides;  therefore  the  angle  DFG  is  equal  to 
the  angle  DFE,  and  the  angle  at  G  to  the  angle  at  E:  but  the 
angle  DFG  is  equal  to  the  angle  ACB;  therefore  the  angle  ACB 
is  equal  to  the  angle  DFE:  and  the  angle  BAG  is  equal  to  the  an- 
gle EDF  (Hyp.);  wherefore  also  the  remaining  angle  at  B  is 
equal  to  the  remaining  angle  at  E.  Therefore  the  triangleABC 
is  equiangular  to  the  triangle  DEF.  Wherefore,  if  two  triangles, 
&c.  Q.  E.  D. 

PROP.  VII.  THEOR. 

If  two  triangles  have  one  angle  of  the  one  equal  to 
one  angle  of  the  other,  and  the  sides  about  two  other 
angles  proportionals,  then,  if  each  of  the  remaining  an- 
gles be  either  less,  or  not  less,  than  a  right  angle;  or 
if  one  of  them  be  a  right  angle;  the  triangles  shall  be 
equiangular,  and  have  those  angles  equal  about  which 
the  sides  are  proportionals.^ 

Let  the  two  triangles  ABC,  DEF  have  one  angle  in  the  one 
equal  to  one  angle  in  the  other,  viz.  the  angle  BAG  to  the  angle 
EDF,  and  the  sides  about  two  other  angles  ABG,  DEF  propor- 
tionals, so  that  AB  is  to  BG,  as  DE  to  EF;  and  in  the  first  case, 
let  each  of  the  remaining  angles  at  G,  F  be  less  than  a  right  an- 
gle. The  triangle  ABG  is  equiangular  to  the  triangle  DEF,  viz. 
the  angle  ABG  is  equal  to  the  angle  DEF,  and  the  remaining  an- 
gle at  C,  to  the  remaining  angle  at  F. 

For,  if  the  angles  ABG,  DEF  be  not  equal,  one  of  them  is 
greater  than  the  other:  let  ABG  be  the  greater,  and  at  the  point 
B,  in  the  straight  line  AB,  make  A 

the  angle  ABG  equal  to  the  an- 
gle (23.  1.)  DEF:  and  because 
the  angle  at  A  is  equal  to  the 
angle  at  D,  and  the  angle  ABG 
to  the  angle  DEF;  the  remain- 
ing angle  AGB  is  equal  (32.  1.) 
to  the  remaining  angle  DFE:  therefore  the  triangle  ABG  is  equi- 
angular to  the  triangle  DEF;  wherefore,  (4.  6.)  as  AB  is  to  BG, 
so  is  DE  to  EF;  but  as  DE  to  EF,  so,  by  hypothesis,  is  AB  to  BG; 
therefore  as  AB  to  BG,  so  is  AB  to  BG  (11.  5.);  and  because  AB 
has  the  same  ratio  to  each  of  the  lines  BG,  BG;  BG  is  equal  (9.  5.) 
to  BG,  and  therefore  the  angle  BGG  is  equal  to  the  angle  BGG 
(5.  1.);  but  the  angle  BGG  is,  by  hypothesis,  less  than  a  right  an- 
gle; therefore  also  the  angle  BGG  is  less  than  a  right  angle,  and 

**  See  Note. 


BOOK  VI.  ir^/if^.       THE    ELEMENTS    OF  EUCLID. 


133 


G 


the  adjacent  angle  AGB  must  be  greater  than  a  right  angle  (13. 1.). 
But  it  was  proved  that  the  angle  AGB  is  equal  to  the  angle  at  F; 
therefore  the  angle  at  F  is  greater  than  a  right  angle:  but  by  the 
hypothesis,  it  is  less  than  a  right  angle  which  is  absurd.  There- 
fore the  angles  ABC,  DEF  are  not  unequal,  that  is,  they  are 
equal:  and  the  angle  at  A  is  equal  to  the  angle  at  Dj  wherefore  the 
remaining  angle  at  C  is  equal  to  the  remaining  angle  at  F:  there- 
fore the  triangle  ABC  is  equiangular  to  the  triangle  DEF. 

Next,  let  each  of  the  angles  at  C,  F,  be  not  less  than  a  right  an- 
gle: the  triangle  ABC  is  also  in  this  case  equiangular  to  the  tri- 
angle DEF. 

The   same  construction  being  A 

made,  it  may  be  proved  in  like 
manner  that  BC  is  equal  to  BG, 
and  the  angle  at  C  equal  to  the 
angle  BGC:  but  the  angle  at  C 
is  not  less  than  a  right  angle; 
therefore  the  angle  BGC   is  not      B  C       E  F 

less  than  a  right  angle;  wherefore  two  angles  of  the  triangle  BGC 
are  together  not  less  than  two  right  angles,  which  is  impossible 
(17.  I.):  and  therefore  the  triangle  ABC  may  be  proved  to  be 
equiangular  to  the  triangle  DEF,  as  in  the  first  case. 

Lastly,  let  one  of  the  angles  at  C,  F,  viz.  the  angle  at  C,  be  a 
right  angle;  in  this  case  likewise  the  triangle  ABC  is  equiangu- 
lar to  the  triangle  DEF. 

For,  if  they  be  not  equiangu-  A 

lar,  make,  at  the  point  B  of  the 
straight  line  AB,  the  angle  ABG 
equal  to  the  angle  DEF;  then 
it  may  be  proved,  as  in  the  first 
case,  that  BG  is  equal  to  BC; 
but  the  angle  BCG  is  a  right 
angle,  therefore  (5.  1.)  the  an- 
gle BGC  is  also  a  right  angle; 
whence  two  of  the  angles  of 
the  triangle  BGC  are  together 
not  less  than  two  right  angles, 
which  is  impossible  (17.  1.); 
therefore  the  triangle  ABC  is 
equiangular  to  the  triangle 
DEF.  Wherefore,  if  two  tri- 
angles, he.     Q.  E.  D. 


B 


B 


PROP.  VIII.    THEOR. 


In  a  right  angled  triangle,  if  a  perpendicular  be 
drawn  from  the  right  angle  to  the  base,  the  triangles 
on  each  side  of  it  are  similar  to  the  whole  triangle,  and 
to  one  another.* 


See  Note. 


134 


THE  ELEMENTS  OF  EUCLID. 


BOOK  VI. 


Let  ABC  be  a  right  angled  triangle,  having  the  right  angle 
BAC|  and  from  the  point  A  let  AD  be  drawn  perpendicular  to 
the  base  BC:  the  triangles  ABD,  ADC  are  similar  to  the  whole 
triangle  ABC,  and  to  one  another. 

Because  the  angle  BAC  is  equal  to  the  angle  ADB,  each  of 
them  being  a  right  angle,  and  that  the  angle  at  B  is  common  to 
the  two  triangles  ABC,  ABD;  the 
remaining  angle  ACB  is  equal  to 
the  remaining  angle  BAD  (32.  1.): 
therefore  the  triangle  ABC  is  equi- 
angular to  the  triangle  ABD,  and 
the  sides  about  their  equal  angles 
are  proportionals  (4.  6.);  wherefore 
the  triangles  are  similar  ( 1 .  Def.  6.); 
in  the  like  manner  it  may  be  demonstrated,  that  the  triangle  ADC 
is  equiangular  and  similar  to  the  triangle  ABC:  and  the  triangles 
ABD,  ADC,  being  both  equiangular  and  similar  to  ABC,  are 
equiangular  and  similar  to  each  other.  Therefore,  in  a  right  an- 
gled, gcc.     Q.  E.  D. 

Cor.  From  this  it  is  manifest,  that  the  perpendicular  drawn 
from  the  right  angle  of  a  right  angled  triangle  to  the  base,  is  a 
mean  proportional  between  the  segments  of  the  base:  and  also 
that  each  of  the  sides  is  a  mean  proportional  between  the  base, 
and  its  segment  adjacent  to  that  side:  because  in  the  triangles 
BDA,  ADC,  BD  is  to  DA  as  DA  to  DC  (4.  6.);  and  in  the  trian- 
gles ABC,  DBA,  BC  is  to  BA,  as  BA  to  BD  (4.  6.);  and  in  the 
triangles  ABC,  ACD,  BC  is  to  CA  as  CA  to  CD  (4.  6.). 


PROP.  IX.  PROB. 

From  a  given  straight  line  to  cut  off  any  part  re- 
quired.* 

Let  AB  be  the  given  straight  line;  it  is  required  to  cut  off  any 
part  from  it. 

From  the  point  A  draw  a  straight  line  AC  making  any  angle 
with  AB;  and  in  AC  take  any  point  D,  and  take  AC  the  same 
multiple  of  AD,  that  AB  is  of  the  part  which  is 
to  be  cut  off  from  it:  join  BC,  and  draw  DE  pa- 
rallel to  it:  then  AE  is  the  part  required  to  be 
cut  off. 

Because  ED  is  parallel  to  one  of  the  sides  of 
the  triangle  ABC,  viz.  to  BC,  as  CD  is  to  DA, 
so  is  (2.  6.)  BE  to  EA;  and,  by  composition  (18. 
5.)  CA  is  to  AD  as  BA  to  AE:  but  CA  is  a  mul- 
tiple of  AD;  therefore  (D.  5.)  BA  is  the  same 
multiple  of  AE:  whatever  part  therefore  AD  is 
of  AC,  AE  is  the  same  part  of  AB:  wherefore, 
from  the  straight  line  AB  the  part  required  is  cut  off. 
was  to  be  done. 

*  See  Note. 


BOOK  VI. 


THE  ELEMENTS  OF  EUCLID. 


135 


PROP.  X.  PROB. 

To  divide  a  given  straight  line  similarly  to  a  given 
divided  straight  line,  that  is,  into  parts  that  shall  have 
the  same  ratios  to  one  another  which  the  parts  of  the 
divided  given  straight  line  have. 

Let  AB  be  the  straight  line  given  to  be  divided,  and  AC  the 
divided  line;  it  is  required  to  divide  AB  similarly  to  AC. 

Let  AC  be  divided  in  the  points  D,  E;  and  let  AB,  AC  be 
placed  so  as  to  contain  any  angle,  and  join  BC,  and  through  the 
points  D,  E  draw  (31.  1.)  DF,  EG  parallels  to  it;  and  through  D 
draw  DHK  parallel  to  AB:  therefore  each  of  the  figures  FH,  HB^ 
is  a  parallelogram;  wherefore  DH  is  equal  (34.  1.)  to  FG,  and 
HK  to  GB:  and  because  HE  is  parallel 
to  KC,  one  of  the  sides  of  the  triangle 
DKC,  as  CE  to  ED,  so  is  (2.  6.)  KH  to 
HD:  but  KH  is  equal  to  BG,  and  HD 
to  GF;  therefore  as  CE  to  ED,  so  is 
BG  to  GF;  again,  because  FD  is  paral- 
lel to  EG,  one  of  the  sides  of  the  trian- 
gle AGE,  as  ED  to  DA,  so  is  GF  to 
FA;  but  it  has  been  proved  that  CE  is 
to  ED  as  BG  to  GF;  and  as  ED  to  DA, 
so  GF  to  FA:  therefore  the  given  straight  line  AB  is  divided  si- 
milarly to  AC.     Which  was  to  be  done. 

PROP.  XL  PROB. 

To  find  a  third  proportional  to  two  given  straight 
lines. 

Let  AB,  AC  be  the  two  given  straight  lines,  and  let  them  be 
placed  so  as  to  contain  any  angle;  it  is  required  to  find  a  third 
proportional  to  AB,  AC. 

Produce  AB,  AC  to  the   points  D,  E:  and        A 
make  BD  equal  to  AC ;  and  having  joined  BC, 
through  D  draw  DE  parallel  to  it  (31.  1.). 

Because  BC  is  parallel  to  DE,  a  side  of  the 
triangle  ADE,  AB  is  (2.  6.)  to  BD,  as  AC  to 
CE:  but  BD  is  equal  to  AC;  as  therefore  AB 
to  AC,  so  is  AC  to  CE.  Wherefore,  to  the  two 
given  straight  lines  AB,  AC  a  third  propor- 
tional CE  is  found.     Which  was  to  be  done. 

PROP.  Xn.  PROB. 

To  find  a  fourth  proportional  to  three  given  straight 
lines. 

Let  A,  B,  C  be  the  three  given  straight  lines;  it  is  required  to 
find  a  fourth  proportional  to  A,  B,  C. 


136  THE  ELEMENTS  OF  EUCLID.  BOOK  VI. 

Take  two  straight  lines  DE,  DF,  containing  any  angle  EDF: 
and  upon  these  make  DG  equal  to  D 

A,  GE  equal  to  B,  and  DH  equal  ^  \ 

to  C;  and  having  joined  GH,  draw 

EF  parallel  (31.  I.)  to  it  through  /     \  S 

the  point  E:  and  because  GH  is 
parallel  to  EF,  one  of  the  sides  of  / 

the  triangle  DEF,  DG  is  to  GE,     ^  / 
as  DH  to  HF  (2.  6.)j  hut  DG  is     ^/ 
equal  to  A,  GE  to  B,  and  DH  to      / 

C;  therefore,  as  A  is  to  B,  so  is  C     ^ —  

to   HF:    wherefore   to    the    three    E  F 

given  straight  lines  A,  B,  C,  a  fourth  proportional  HF  is  found. 

Which  was  to  be  done. 


PROP.  XHI.  PROB. 

To  find  a  mean  proportional   between  two  given 
straight  lines. 

Let  AB,  BC  be  the  two  given  straight  linesj  it  is  required   to 
find  a  mean  proportional  between  them. 

Place  AB,  BC  in  a  straight  line,  and  upon  AC  describe  the 
semicircle  ADC,  and  from  the  point 
B  draw  (11.  1.)  BD  at  right  angles 
to  AC,  and  join  AD,  DC. 

Because  the  angle  ADC  in  a  semi- 
circle is  a  right  angle  (31.  3.),  and 
because  in  the  right  angled  triangle 
ADC,  DB  is  drawn  from  the  right  a" 
angle  perpendicular  to  the  base,  DB 
is  a  mean  proportional  between  AB,  BC,  the  segments  of  the  base 
(Cor.  8.  6.):  therefore  between  the  two  given  straight  lines  AB,  BC 
a  mean  proportional  DB  is  found.     Which  was  to  be  done. 


PROP.  XIV.  THEOR. 

Equal  parallelograms  which  have  one  angle  of  the 
one  equal  to  one  angle  of  the  other,  have  their  sides 
about  the  equal  angles  reciprocally  proportional:  and 
parallelograms  that  have  one  angle  of  the  one  equal  to 
one  angle  of  the  other,  and  their  sides  about  the  equal 
angles  reciprocally  proportional,  are  equal  to  one  ano- 
ther. 

Let  AB,  BC  be  equal  parallelograms,  which  have  the  angles  at 
B  equal,  and  let  the  sides  DB,  BE  be  placed  in  the  same  straight 
line:  wherefore  also  FB,  BG  are  in  one  straight  line  (14.  1.):  the 
sides  of  the  parallelograms  AB,  BC,  about  the  equal  angles,  are 
reciprocally  proportional,  that  is,  DB  is  to  BE,  as  GB  to  BF. 


BOOK  VI.  THE  ELEMENTS  OF  EUCLID.  137 

Complete  the  parallelogram  FE:  and  because  the  parallelogram 
AB  is  equal  to  BC,  and  that  FE  is     A  F 

another  parallelogram,  AB  is  to  FE,      T 
as  BC  to  FE  (7.  5.):  but  as  AB  to       \ 
FE,  so  is  the  base  DB  to  BE  (1.6.);        ^— 
and  as  BC  to  FE,  so  is  the  base  GB  d 

to  BF:  therefore  as  DB  to  BE,  so  is 
GB  to  BF  (11.  5.).  Wherefore  the 
sides  of  the  parallelograms  AB,  BC 
about  their  equal  angles  are  reci- 
procally proportional. 

But,  let  the  sides  about  the  equal  angles  be  reciprocally  propor- 
tional, viz.  as  DB  to  BE,  so  GB  to  BF;  the  parallelogram  AB  is 
equal  to  the  parallelogram  BC. 

Because  as  DB  to  BE,  so  is  GB  to  BF;  and  as  DB  to  BE,  so 
is  the  parallelogram  AB  to  the  parallelogram  FE;  and  as  GB  to 
BF,  so  is  the  parallelogram  BC  to  the  parallelogram  FE;  there- 
fore as  AB  to  FE,  so  BC  to  FE  (9.  5.):  wherefore  the  parallelo- 
gram AB  is  equal  (9.  5.)  to  the  parallelogram  BC.  Therefore, 
equal  parallelograms,  Sec.  Q.  E.  D. 

PROP.  XV.  THEOR. 
EquAL  triangles,  which  have  one  angle  of  the  one 
equal  to  one  angle  of  the  other,  have  their  sides  about 
the  equal  angles  reciprocally  proportional ;  and  triangles 
which  have  one  angle  in  the  one  equal  to  one  angle  in 
the  other,  and  their  sides  about  the  equal  angles  re- 
ciprocally proportional,  are  equal  to  one  another. 

Let  ABC,  ADE  be  equal  triangles,  which  have  the  angle  BAC 
equal  to  the  angle  DAE;  the  sides  about  the  equal  angles  of  the 
triangles  are  reciprocally  proportional;  that  is,  CA  is  to  AD,  as 
EA  to  AB. 

Let  the  triangles  be  placed  so  that  their  sides  CA,  AD  be  in 
one  straight  line;  wherefore  also  EA  and  AB  are  in  one  straight 
line  (14.  1.)  and  join  BD.  Because  the  triangle  ABC  is  equal 
to  the  triangle  ADE,  and  that  ABD  B  D 

is  another  triangle,  therefore  as  the 
triangle  CAB  is  to  the  triangle 
BAD,  so  is  triangle  EAD  to  tri- 
angle DAB  (7.  5.):  but  as  triangle 
CAB  to  triangle  BAD,  so  is  the 
base  CA  to  AD  (l.  6.);  and  as  tri- 
angle EAD  to  triangle  DAB,  so  is 

the  base  EA  to  AB  (1.  6.):  as  there-       C  E 

fore  CA  to  AD,  so  is  EA  to  AB  (l  1.  5.);  wherefore  the  sides  of 
the  triangles  ABC,  ADE  about  the  equal  angles  are  reciprocally 
proportional. 

But  let  the  sides  of  the  triangles  ABC,  ADE  about  the  equal 
angles  be  reciprocally  proportional,  viz.  CA  to  AD,  as  EA  to  AB; 
the  triangle  ABC  is  equal  to  the  triangle  ADE. 


138  THE  ELEMENTS  OF  EUCLID.  BOOK  VI., 

Having  joined  BD  as  before;  because  as  CA  to  AD  so  is  EA 
to  AB;  and  as  CA  to  AD,  so  is  *  "angle  BAG  to  triangle  BAD 
(l.  6.);  and  as  EA  to  AB,  so  is  triangle  EAD  to  triangle  BAD 
(1.  6.);  therefore  (11.  5.)  as  triangle  BAG  to  triangle  BAD,  so  is 
triangle  EAD  to  triangle  BAD;  that  is,  the  triangles  BAG,  EAD 
have  the  same  ratio  to  the  triangle  BAD;  wherefore  the  triangle 
ABG  is  equal  (9.  5.)  to  the  triangle  ADE.  Therefore,  equal  tri- 
angles, 8cc.  Q.  E.  D. 

PROP.  XVI.  THEOR. 

If  four  straight  lines  be  proportionals,  the  rectangle 
contained  by  the  extremes  is  equal  to  the  rectangle 
contained  by  the  means:  and  if  the  rectangle  contained 
by  the  extremes  be  equal  to  the  rectangle  contained  by 
the  means,  the  four  straight  lines  are  proportionals. 

Lei  the  four  straight  lines  AB,  GD,  E,  F,  be  proportionals,  viz- 
as  AB  to  GD,  so  is  E  to  F;  the  rectangle  contained  by  AB,  F  is 
equal  to  the  rectangle  contained  by  GD,  E. 

From  the  points  A,  G  draw  (11.  1.)  AG,  GH  at  right  angles  to 
AB,  GD;  and  make  AG  equal  to  F,  and  GH  equal  to  E,  and 
complete  the  parallelograms  BG,  DH;  because  as  AB  to  GD,  so 
is  E  to  F;  and  that  E  is  equal  to  GH,  and  F  to  AG;  AB  is  (7.  5.) 
to  GD,  as  GH  to  AG:  therefore  the  sides  of  the  parallelograms 
BG,  DH  about  the  equal  angles  are  reciprocally  proportional; 
but  parallelograms  which  have  their  sides  about  equal  angles  re- 
ciprocally proportional,  are  equal  to  one  another  (14.  6.);  there- 
fore the  parallelogram  BG  is  equal  to  the  parallelogram  DH, 

and  the  parallelogram  BG  is  con-   E 

tained  by  the  straight  lines  AB, 
F,  because  AG  is  equal  to  F;  and 
the  parallelogram  DH  is  contain- 
ed by  GD  and  E,  because  GH  is 
equal  to  E;  therefore  the  rectangle 
contained  by  the  straight  lines  AB, 
F  is  equal  to  that  v/hich  is  con- 
tained by  CD  and  E. 

And  if  the  rectangle  contained 
by  the  straight   lines  AB,  F  be 
equal  to  that  which  is  contained  by  GD,  E;  these  four  lines  are 
proportionals,  viz.  AB  is  to  GD,  as  E  to  F. 

The  same  construction  being  made,  because  the  rectangle  con- 
tained by  the  straight  lines  AB,  F  is  equal  to  that  which  is  con- 
tained by  GD,  E,  and  that  the  rectangle  BG  is  contained  by  AB, 
F,  because  AG  is  equal  to  F;  and  the  rectangle  DH  by  GD,  E, 
because  GH  is  equal  to  E;  therefore  the  parallelogram  BG  is 
equal  to  the  parallelogram  DH,  and  they  are  equiangular:  but  the 
sides  about  the  equal  angles  of  equal  parallelograms  are  recipro- 
cally proportional  (14.  6.);  wherefore,  as  AB  to  GD,  so  is  GH  to 
AG;  and  GH  is  equal  to  E,  and  AG  to  F:  as  therefore  AB  is  to 
CD,  so  E  to  F.     Wherefore,  if  four,  8cc.  Q.  E.  D. 


BOOK  VI.  THE  ELEMENTS  OF  EUCLID.  139 

PROP.  XVII.  THEOR. 

If  three  straight  lines  be  proportionals,  the  rectangle 
contained  by  the  extremes  is  equal  to  the  square  of  the 
mean:  and  if  the  rectangle  contained  by  the  extremes 
be  equal  to  the  square  of  the  mean,  the  three  straight 
lines  are  proportionals. 

Let  the  three  straight  lines  A,  B,  C  be  proportionals,  viz.  as  A 
to  B,  so  B  to  C;  the  rectangle  contained  by  A,  C  is  equal  to  the 
square  of  B. 

Take  D  equal  to  B;  and  because  as  A  to  B,  so  B  to  C,  and  that 
B  is  equal  to  Dj  A  is  (7.  5.)  to  B,  as  D  to  C;  but  if  four  straight 

lines   be   proportionals,   the  A 

rectangle  contained  by  the  ex-  B 

tremes  is  equal  to  that  which  D 

is  contained  by  the  means  (16.  C 

6.) :   therefore  the  rectangle 

contained  by  A,  C  is  equal  to       r  ]  C  D 

that  contained  iDy  B,  D.    But 

the  rectangle  contained  by  B,       I 

D  is  the  square  of  B  ;    be-  A  B 

cause  B  is  equal  to  D;  therefore  the  rectangle  contained  by  A,  C 
is  equal  to  the  square  of  B. 

And  if  the  rectangle  contained  by  A,  C  be  equal  to  the  square 
of  B;  A  is  to  B,  as  B  to  C. 

The  same  construction  being  made,  because  the  rectangle  con- 
tained by  A,  C  is  equal  to  the  square  of  B,  and  the  square  of  B 
is  equal  to  the  rectangle  contained  by  B,  D,  because  B  is  equal  to 
D;  therefore  the  rectangle  contained  by  A,  C  is  equal  to  that 
contained  by  B,  D:  but  if  the  rectangle  contained  by  the  ex- 
tremes be  equal  to  that  contained  by  the  means,  the  four  straight 
lines  are  proportionals  (16.  6.);  therefore  A  is  to  B,  as  D  to  C; 
but  B  is  equal  to  D;  wherefore  as  A  to  B,  so  B  to  C.  Therefore, 
if  three  straight  lines,  Sec.  Q.  E.  D. 

PROP.  XVIII.  PROB. 

Upon  a  given  straight  line  to  describe  a  rectilineal 
figure  similar  and  similarly  situated  to  a  given  rectili- 
neal figure.* 

Let  AB  be  the  given  straight  line,  and  CDEF  the  given  recti- 
lineal figure  of  four  sides;  it  is  required  upon  the  given  straight 
line  AB  to  describe  a  rectilineal  figure  similar  and  similarly  situ- 
ated to  CDEF. 

Join  DF,  and  at  the  points  A,  B,  in  the  straight  line  AB,  make 
(23.  1.)  the  angle  BAG  equal  to  the  angle  at  C,  and  the  angle  ABG 
equal  to  the  angle  CDF;  therefore  the  remaining  angle  CFD  is 
equal  to  the  remaining  angle  AGB  (32.  1.);  wherefore  the  triangle 

*  See  Note. 


140  THE  ELEMENTS  OF  EUCLID.  BOOK  VI. 

FCD  is  equiangular  to 
the  triangle  GAB:  again 
at  the  points  G,  B,  in  the 
straight  line  GB,  make 
(23.  I.)  the  angle  BGH 
equal  to  the  angle  DFE, 
and  the  angle  GBH  equal 
to   FDE:   therefore   the     A  BCD 

remaining  anj^le  FED  is  equal  to  the  remaining  angle  GHB5  and 
the  triangle  FDE  equiangular  to  the  triangle  GBH:  then,  because 
the  angle  AGB  is  equal  to  the  angle  CFD,  and  BGH  to  DFE,  the 
whole  angle  AGH  is  equal  to  the  whole  CFE:  for  the  same  rea- 
son, the  angle  ABH  is  equal  to  the  angle  CDE|  also  the  angle  at 
A  is  equal  to  the  angle  at  C,  and  the  angle  GHB  to  FED;  there- 
fore the  rectilineal  figure  ABHG  is  equiangular  to  CDEF:  but 
likewise  these  figures  have  their  sides  about  the  equal  angles 
proportionals;  because  the  triangles  GAB,  FCD  being  equian- 
gular, BA  is  (4.  6.)  to  AG,  as  DC  to  CF;  and  because  AG  is  to 
GB,  as  CF  to  FD;  and  as  GB  to  GH,  so,  by  reason  of  the  equi- 
angular triangles  13 GH,  DFE,  is  FD  to  FE;  therefore,  ex  sequali 
(22.  5.),  AG  is  to  GH,  as  CF  to  FE:  in  the  same  manner  it  may 
be  proved  that  AB  is  to  BH,  as  CD  to  DE:  and  GH  is  to  HB,  as 
FE  to  ED  (4.  6.).  Wherefore,  because  the  rectilineal  figures 
ABHG,  CDEF  are  equiangular,  and  have  their  sides  about  the 
equal  angles  proportionals,  they  are  similar  to  one  another 
(1.  def.  6.). 

Next,  let  it  be  required  to  describe  upon  a  given  straight  line 
AB,  a  rectilineal  figure  similar  and  similarly  situated  to  the  rec- 
tilineal figure  CDKEF. 

Join  DE,  and  upon  the  given  straight  line  AB  describe  the 
rectilineal  figure  ABHG  similar  and  similarly  situated  to  the 
quadrilateral  figure  CDEF,  by  the  former  case;  and  at  the  points 
B,  H,  in  the  straight  line  BH,  make  the  angle  HBL  equal  to  the 
angle  EDK,  and  the  angle  BHL  equal  to  the  angle  DEK;  there- 
fore the  remaining  angle  at  K  is  equal  to  the  remaining  angle  at 
L:  and  because  the  figures  ABHG,  CDEF  are  similar,  the  angle 
GHB  is  equal  to  the  angle  FED,  and  BHL  is  equal  to  DEK^ 
wherefore  the  whole  angle  GHL  is  equal  to  the  whole  angle  FEK: 
for  the  same  reason  the  angle  ABL  is  equal  to  the  angle  CDK; 
therefore  the  five  sided  figures  AGHLB,  CFEKD  are  equiangular; 
and  because  the  figures  AGHB,  CFED  are  similar,  GH  is  to  HB, 
as  FE  to  ED;  and  as  HB  to  HL,  so  is  ED  to  EK  (4.  6.);  therefore, 
ex  eequali  (22.  5.),  GH  is  to  HL,  as  FE  to  EK:  for  the  same  rea- 
son, AB  is  to  BL,  as  CD  to  DK:  and  BL  is  to  LH,  as  (4.  6.)  DK 
to  KE;  because  the  triangles  BLH,  DKE  are  equiangular;  there- 
fore, because  the  five  sided  figures  AGHLB,  CFEKD  are  equi- 
angular, and  have  their  sides  about  the  equal  angles  proportionals, 
they  are  similar  to  one  another;  and  in  the  same  manner  a  recti- 
lineal figure  of  six  or  more  sides  may  be  described  upon  a  given 
straight  line  similar  to  one  given,  and  so  on.  Which  was  to  be 
done. 


BOOK  VI.  THE  ELEMENTS  OF  EUCLID.  141 


PROP.  XIX.  THEOR. 

Similar  triangles  are  to  one  another  in  the  duphcate 
ratio  of  their  homologous  sides. 

Let  ABC,  DEF  be  similar  triangles,  having  the  angle  B  equal 
to  the  angle  E,  and  let  AB  be  to  BC,  as  DE  to  EF,  so  that  the 
side  BC  is  homologous  to  EF  (12.  def.  5.);  the  triangle  ABC  has 
to  the  triangle  DEF  the  duplicate  ratio  of  that  which  BC  has  to 
EF. 

fake  BG  a  third  proportional  to  BC,  EF  (ll.  6.),  so  that  BC 
is  to  EF,  as  EF  to  BG,  and  join  GA;  then,  because  as  AB  to  BC, 
so  DE  to  EF,  alternately  (16.  5.),  AB  is  to  DE,  as  BC  to  EF:  but 
as  BC  to  EF;  so  is  EF  to  BG;  therefore  (11.  5.),  as  AB  to  DE, 
so  is  EF  to  BG;  wherefore  the  sides  of  the  triangles  ABG,  DEF 
which  are  about  the  equal  angles,  are  reciprocally  proportional: 
but  triangles  which  have  the  sides  about  two  equal  angles  recipro- 
cally proportional,  are  A 
equal  to  one  another  (15.  w\  . 
6.):  therefore  the  trian-  -  ^ 
gle  ABG  is  equal  to  the 
triangle  DEF:  and  be- 
cause as  BC  is  to  EF,  so 
EF  to  BG;  and  that  if 
three  straight  lines  be 
proportionals,  the  first  is 
is  said  (lO.  def.  5.)  to  have  to  the  third  the  duplicate  ratio  of  that 
which  it  has  to  the  second;  BC  therefore  has  to  BG  the  duplicate 
ratio  of  that  which  BC  has  to  EF:  but  as  BC  to  BG,  so  is  (l.  6.) 
the  triangle  ABC  to  the  triangle  ABG.  Therefore  the  triangle 
ABC  has  to  the  triangle  ABG  the  duplicate  ratio  of  that  which 
BC  has  to  EF;  but  the  triangle  ABG  is  equal  to  the  triangle 
DEF;  wherefore  also  the  triangle  ABC  has  to  the  triangle  DEF 
the  duplicate  ratio  of  that  which  BC  has  to  EF.  Therefore,  si- 
milar triangles.  Sec.  Q.  E.  D. 

Cor.  From  this  it  is  manifest,  that  if  three  straight  lines  be 
proportionals,  as  the  first  is  to  the  third,  so  is  any  triangle  upon 
the  first  to  a  similar  and  similarly  described  triangle  upon  the 
second. 

PROP.  XX.  THEOR. 

Similar  polygons  may  be  divided  into  the  same  num- 
ber of  similar  triangles,  having  the  same  ratio  to  one 
another  that  the  polygons  have :  and  the  polygons  have 
to  one  another  the  duplicate  ratio  of  that  which  their 
homologous  sides  have. 

Let  ABCDE,  FGHKL  be  similar  polygons,  and  let  AB  be  the 
homologous  side  to  FG:  the  polygons  ABCDE,  FGHKL  may  be 
divided  into  the  same  number  of  similar  triangles,  whereof  each 
to  each  has  the  same  ratio  which  the  polygons  have;  and  the  po- 


142  THE  ELEMENTS  OF  EUCLID.  BOOK  VI. 

lygon  ABCDE  has  to  the  polygon  FGHKL  the  duplicate  ratio  of 
that  which  the  side  AB  has  to  the  side  FG. 

Join  BE,  EC,  GL,  LH:  and  because  the  polygon  ABCDE  is 
similar  to  the  polygon  FGHKL,  the  angle  BAE  is  equal  to  the 
angle  GFL  (l.  def.  6.), and  BA  is  to  AE,  as  GF  to  FL  (1.  def.  6.)j 
wherefore  because  the  triangles  ABE,  FGL,  have  an  angle  in  one 
equal  to  an  angle  in  the  other,  and  their  sides  about  these  equal 
angles  proportionals^  the  triangle  ABE  is  equiangular  (6.  6.),  and 
therefore  similar  to  the  triangle  FGL  (4.  6.);  wherefore  the  angle 
ABE  is  equal  to  the  angle  FGL:  and,  because  the  polygons  are 
similar,  the  whole  angle  ABC  is  equal  (l.  def.  6.)  to  the  whole 
angle  FGH;  therefore  the  remaining  angle  EBC  is  equal  to  the 
remaining  angle  LGH:  and  because  the  triangles  ABE,  FGL  are 
similar,  EB  is  to  BA,  as  LG  to  GF  (1.  def.  6.);  and  also,  because 
the  polygons  are  similar,  AB  is  to  BC,  as  FG  to  GH  (l.  def.  6.); 
therefore,  ex  aequali  (22.  5.),  EB  is  to  BC,  as  LG  to  GHj  that  is, 
the  sides  about  the  equal  angles  EBC,  LGH  are  proportionals^ 
therefore  (22.  5.) 
the  triangle  EBC 
is  equiangular  to 
the  triangle  LGH, 
and  similar  to  it 
(4.  6.).  For  the 
same  reason,  the 
triangle  EC  D like- 
wise is  similar  to  -^ 
the  triangle  LKH;  therefore  the  similar  polygons  ABCDE, 
FGHKL  are  divided  into  the  same  number  of  similar  triangles. 

Also  these  triangles  have,  each  to  each,  the  same  ratio  which 
the  polygons  have  to  one  another,  the  antecedents  being  ABE, 
EBC,  ECD,  and  the  consequents  FGL,  LGH,  LHK:  and  the  po- 
lygon ABCDE  has  to  the  polygon  FGHKL  the  duplicate  ratio  of 
that  which  the  side  AB  has  to  the  homologous  side  FG. 

Because  the  triangle  ABE  is  similar  to  the  triangle  FGL,  ABE 
has  to  FGL  the  duplicate  ratio  (19.  6.)  of  that  which  the  side  BE 
has  to  the  side  GL;  for  the  same  reason,  the  triangle  BEC  has  to 
GLH  the  duplicate  ratio  of  that  which  BE  has  to  GL;  therefore, 
as  the  triangle  ABE  to  the  triangle  FGL,  so  (l  1.  5.)  is  the  trian- 
gle BEC  to  the  triangle  GLH.  Again,  because  the  triangle  EBC 
is  similar  to  the  triangle  LGH,  EBC  has  to  LGH  the  duplicate 
ratio  of  that  which  the  side  EC  has  to  the  side  LH:  for  the  same 
reason,  the  triangle  ECD  has  to  the  triangle  LHK,  the  duplicate 
ratio  of  that  which  EC  has  to  LH;  as  therefore  the  triangle  EBC 
to  the  triangle  LGH,  so  is  (11.  5.)  the  triangle  ECD  to  the  trian- 
gle LHK;  but  it  has  been  proved  that  the  triangle  EBC  is  like- 
wise to  the  triangle  LGH,  as  the  triangle  ABE  to  the  triangle 
FGL.  Therefore  as  the  triangle  ABE  is  to  the  triangle  FGL,  so 
is  triangle  EBC  to  triangle  LGH,  and  triangle  ECD  to  triangle 
LHK:  and  therefore  as  one  of  the  antecedents  to  one  of  the  con- 
sequents? so  are  all  the  antecedents  to  all  the  consequents  (12.  5.). 
Wherefore  as  the  triangle  ABE  to  the  triangle  FGL,  so  is  the 
polygon  ABCDE  to  the  polygon  FGHKL;  but  the  triangle  ABE 


BOOK  VI.  THE  ELEMENTS  OF  EUCLID.  143 

has  to  the  triangle  FGL,  the  duplicate  ratio  of  that  which  the  side 
AB  has  to  the  homologous  side  FG.  Therefore  also  the  polygon 
ABCDE  has  to  the  polygon  FGHKL  the  duplicate  ratio  of  that 
which  AB  has  to  the  homologous  side  FG.  Wherefore,  similar 
polygons,  Sec.  Q.  E.  D. 

Cor.  1.  In  like  manner,  it  may  be  proved,  that  similar  four 
sided  figures,  or  of  any  number  of  sides,  are  one  to  another  in  the 
duplicate  ratio  of  their  homologous  sides,  and  it  has  already  been 
proved  in  triangles.  Therefore,  universally,  similar  rectilineal 
hgures  are  to  one  another  in  the  duplicate  ratio  of  their  homolo- 
gous sides. 

Cor.  2.  And  if  to  AB,  FG,  two  of  the  homologous  sides,  a 
third  proportional  M  be  taken,  AB  has  (10.  def.  5.)  to  M  the  du- 
plicate ratio  of  that  which  AB  has  to  FG:  but  the  four  sided  figure 
or  polygon  upon  AB,  has  to  the  four  sided  figure  or  polygon  upon 
FG  likewise  the  duplicate  ratio  of  that  which  AB  has  to  FG^ 
therefore,  as  AB  is  to  M,  so  is  the  figure  upon  AB  to  the  figure 
upon  FG,  which  was  also  proved  in  triangles  (Cor.  19.  6.).  There- 
fore, universally,  it  is  manifest,  that  if  three  straight  lines  be  pro- 
portionals, as  the  first  is  to  the  third,  so  is  any  rectilineal  figure 
upon  the  first,  to  a  similar  and  similarly  described  rectilineal  figure 
upon  the  second. 

PROP.  XXI.    THEOR. 

Rectilineal  figures  which  are  similar  to  the  same 
rectihneal  figures,  are  also  similar  to  one  another. 

Let  each  of  the  rectilineal  figures.  A,  B  be  similar  to  the  recti- 
lineal figure  C;  the  figure  A  is  similar  to  the  figure  B. 

Because  A  is  similar  to  C;  they  are  equiangular,  and  also  have 
their  sides  about  the  equal  angles  proportionals  (l  def.  6.).  Again, 
because  B  is  similar  to  C, 
they  are  equiangular,   and 

have  their  sides  about  the  y^  Nv         X  C 

equal  angles  proportionals 
(1  def.  6.):  therefore  the 
figures  A,  B  are  each  of 
them  equiangul'ar  to  C,  and 
have  the  sides  about  the  equal  angles  of  each  of  them  and  of  C 
proportionals.  Wherefore  the  rectilineal  figures  A  and  B  are 
equiangular  (1.  Ax.  1.),  and  have  their  sides  about  the  equal  an- 
gles proportionals  (11.  5.).  Therefore  A  is  similar  (l.  def.  6.)  to 
B.  Q.  E.  D. 

PROP.  XXII.  THEOR. 

If  four  straight  lines  be  proportionals,  the  similar 
rectilineal  figures  similarly  described  upon  them  shall 
also  be  proportionals;  and  if  the  similar  rectilineal 
figures  similarly  described  upon  four  straight  lines  be 


144  THE  ELEMENTS  OF  EUCLID.  BOOK  VI. 

proportionals,  those  straight  Unes  shall  be  proportion- 
als. 

Let  the  four  straight  lines  AB,  CD,  EF,  GH  be  proportionals, 
viz.  AB  to  CD,  as  EF  to  GH,  and  upon  AB,  CD  let  the  similar 
rectilineal  figures  KAB,  LCD  be  similarly  described;  and  upon 
EF.  GH  the  similar  rectilineal  figures  MF,  NH  in  like  manner: 
the' rectilineal  figure  KAB  is  to  LCD,  as  MF  to  NH. 

To  AB,  CD  take  a  third  proportional  (U.  6.)  X;  and  to  EF, 
GH  a  third  proportional  O:  and  because  AB  is  to  CD,  as  EF  to 
GH,  and  that  CD  is  (11.  5.)  to  X,  as  GH  to  O;  wherefore,  ex 
cequali  (22.  5.),  as  AB  to  X,  so  is  EF  to  O:  but  as  AB  to  X,  so  is 
(2  Cor.  20.  6.)  the  rectilineal  KAB  to  the  rectilineal  LCD,  and  as 
EF  to  O,  so  is  (2  Cor.  20.  6.)  the  rectilineal  MF  to  the  rectilineal 
NH:  therefore,  as  KAB  to  LCD,  so  (ll.  5.)  is  MF  to  NH. 

And  if  the  rectilineal  KAB  be  to  LCD,  as  MF  to  NH;  the 
straight  line  AB  is  to  CD,  as  EF  to  GH. 

Make  (12.  6.)  as  A B  to  CD  so  EF  to  PR,  and  upon  PR  describe 
(18.  6.)  the  rectilineal  figure  SR  similar  and  similarly  situated  to 

K  L 


M 


E  F     G  H  PR 

either  of  the  figures  MF,  NH:  then,  because  as  AB  to  CD,  so  is 
EF  to  PR,  and  that  upon  AB,  CD  are  described  the  similar  and 
similarly  situated  rectilineals  KAB,  LCD,  and  upon  EF,  PR,  in 
like  manner,  the  similar  rectilineals  MF,  SR;  KAB  is  to  LCD, 
as  MF  to  SR;  but  by  the  hypothesis,  KAB  is  to  LCD,  as  MF 
to  NH:  and  therefore  the  rectilineal  MF  having  the  same  ratio 
to  each  of  the  two  NH,  SR,  these  are  equal  (9.  5.)  to  one  ano- 
ther: they  are  also  similar,  and  similarly  situated;  therefore  GH 
is  equal  to  PR:  and  because  as  AB  to  CD,  so  is  EF  to  PR, 
and  that  PR  is  equal  to  GH;  AB  is  to  CD,  as  EF  to  GH.  If, 
therefore,  four  straight  lines.  Sec.  Q.  E,  D. 

PROP.  XXHL  THEOR. 

Equiangular  parallelograms  have  to  one  another 
the  ratio  which  is  compounded  of  the  ratios  of  their 
sides.^ 

Let  AC,  CF  be  equiangular  parallelograms,  having  the  angle 


» 


See  Note. 


BOOK  VI. 


THE  ELEMENTS  OF  EUCLID. 


145 


B 


BCD  equal  to  the  angle  ECG:  the  ratio  of  the  parallelogram  AC 
to  the  parallelogram  CF,  is  the  same  with  the  ratio  which  is 
compounded  of  the  ratios  of  their  sides. 

Let  BG,  CO,  be  placed  in  a  straight  line;  therefore  DC  and 
CE  are  also  in  a  straight  line  (14.  1,);  and  complete  the  parallel- 
ogram DGj  and,  taking  any  straight  line  K  make  (I2.  6.)  as  BC 
to  CG,  so  K  to  L;  and  as  DC  to  CE,  so  make  (12.  6.)  L  to  M: 
therefore  the  ratios  of  K  to  L,  and  L  to  M,  are  the  same  with  the 
ratios  of  the  sides,  viz.  of  BC  to  CG,  and  DC  to  CE.  But  the 
ratio  of  K  to  M  is  that  which  is  said  to  be  compounded  (A.  def.  5.) 
of  the  ratios  of  K  to  L,  and  L  to  M:  Wherefore  also  K  has  to  M 
the  ratio  compounded  of  the  ratios  of        A  D  H 

the  sides;  and  because  as  BC  to  CG,  so 
is  the  parallelogram  AC  to  the  parallel- 
ogram CH  (1.  6.);  but  as  BC  to  CG,  so 
is  K  to  h;  therefore  K  is  (11.  5.)  to  L,  as 
the  parallelogram.  AC  to  the  parallelo- 
gram GH:  again,  because  as  DC  to  CE, 
so  is  the  parallelogram  CH  to  the  paral- 
lelogram CF;  but  as  DC  to  CE,  so  isL 
to  M;  wherefore  L  is  (l  1.5.)  to  M,  as  the 
parallelogram  CH  to  the  parallelogram 
CF:  therefore  since  it  has  been  proved, 
that  as  K  to  L,  so  is  the  parallelogram 
AC  to  the  parallelogram  CH;  and  as  L  to  M,  so  the  parallelogram 
CH  to  the  parallelogramCF;  ex  asquall  (22.  5.),  K  is  to  M,  as  the 
parallelogram  AC  to  the  parallelogram  CF:  but  K  has  to  M  the 
ratio  which  is  compounded  of  the  ratios  of  the  sides;  therefore 
also  the  parallelogram  AC  has  to  the  parallelogram  CF  the  ratio 
which  is  compounded  of  the  ratios  of  the  sides.  Wherefore,  equi- 
angular parallelograms,  &c.  Q.  E.  D. 


KL  M 


PROP.  XXIV.  THEOR. 

The  parallelograms  about  the  diameter  of  any  paral- 
lelogram are  similar  to  the  whole,  and  to  one  another."^ 

Let  ABCD  be  a  parallelogram,  of  which  the  diameter  is  AC; 
and  EG,  HK  the  parallelograms  about  the  diameter:  the  parallel- 
ograms EG,  HK  are  similar  both  to  the  whole  parallelogram 
ABCD,  and  to  one  another. 

Because  DC,  GF  are  parallels,  the  angle  ADC  is  equal  (29.  1.) 
to  the  angle  AGF:  for  the  same  reason,  because  BC,  EF  are  pa- 
rallels, the  angle  ABC  is  equal  to  the  angle  AEF:  and  each  of  the 
angles  BCD,  EFG  is  equal  to  the  opposite  angle  DAB  (34.  1.), 
and  therefore  are  equal  to  one  another;  wherefore  the  parallelo- 
grams ABCD,  AEFG  are  equiangular;  and  because  the  angle 
ABC  is  equal  to  the  angle  AEF;  and  the  angle  BAG  common  to 
the  two  triangles  BAC,  EAF,  they  are  equiangular  to  one  another; 
therefore  (4.  6.)  as  AB  to  BC,  so  is  AE  to  EF:  and  because  the 

•^  See  Note.  C^m 

T 


J  J 


146 


THE  ELEMENTS  OF  EUULID. 


BOOK  VS. 


oppoiite  sides  of  parallelograms  are 
equal  to  one  another  (34.  1.),  AB  is 
(7.  5.)  to  AD,  as  AE  to  AG;  and  DC 

to  CB,  as  GF  to  FE;  and  also  CD  to     G  !■ ^=4 — /   H 

DA,  as  FG  to  GA:  therefore  the  sides 
of  the  parallelograms  ABCD,  AEFG 
about  the  equal  angles  are  propor- 
tionals; and  they  are  therefore  simi- 
lar to  one  another  (l .  def.  6.):  for  the 
same  reason,  the  parallelogram  ABCD  is  similar  to  the  parallelo- 
gram FHCK.  Wherefore  each  of  the  parallelograms  GE,  KH  is 
similar  to  DB;  but  rectilineal  figures  which  are  similar  to  the 
same  rectilineal  figure  are  also  similar  to  one  another  (21.  6.); 
therefore  the  parallelogram  GE  is  similar  to  KH.  Wherefore, 
the  parallelograms,  Sec.  Q.  E.  D. 

PROP.  XXV.    PROB. 

To  describe  a  rectilineal  figure  which  shall  be  simi- 
lar to  one,  and  equal  to  another  given  rectilineal  figure.* 

Let  ABC  be  the  given  rectilineal  figure,  to  which  the  figure  to 
be  described  is  required  to  be  similar,  and  D  that  to  which  it 
must  be  equal.  It  is  required  to  describe  a  rectilineal  figure  simi- 
lar to  ABC,  and  equal  to  D. 

Upon  the  straight  line  BC  describe  (Cor.  45.  1.)  the  parallelo- 
gram BE  equal  to  the  figure  ABC;  also  upon  CE  describe  (Cor. 
45.  1.)  the  parallelogram  CM  equal  to  D,  and  having  the  angle 
FCE  equal  to  the  angle  CBL;  therefore  BC  and  CF  are  in  a 
straight  line  (29.  1.  14.  1.),  as  also  LE  and  EM:  between  BC  and 
CF  find  (13.  6.)  a  mean  proportional  GH,  and  upon  GH  describe 
(18.  6.)  the  rectilineal  figure  KGH  similar  and  similarly  situated 
to  the  figure  ABC:  and  because  BC  is  to  GH  as  GH  to  CF,  and 
if  three  straight  lines  be  proportionals,  as  the  first  is  to  the  third, 
so  is  (2.  Cor.  20.  6.)  the  figure  upon  the  first  to  the  similar  and 
similarly  described  figure  upon  the  second;  therefore  as  BC  to  CF, 
so  is  the  rectilineal  figure  ABC  to  KGH;  but  as  BC  to  CF,  so  is 
(1.  6.)  the  parallelogram  BE  to  the  parallelogram  EF:  therefore 
as  the  rectilineal  figure  ABC  is  to  KGH,  so  is  the  parallelogram 
BE  to  the  parallelogram  EF  (11.5.):  and  the  rectilineal  figure 

A 

K 


B 


ABC  is  equal  to  the  parallelogram.  BE;  therefore  the  rectilineal 

,— ^  *  See  Note. 


nooK  VI. 


THE  ELEMENTS  OF  EUCLID. 


147 


figure  KGH  is  equal  (14.  5.)  to  the  parallelogram  EF:  but  EF  is 
equal  to  the  figure  D;  wherefore  also  KGH  is  equal  to  D;  and  it 
is  similar  to  ABC.  Therefore  the  rectilineal  figure  KGH  has 
been  described  similar  to  the  figure  ABC,  and  equal  to  D.  Which 
was  to  be  done. 


PROP.  XXVI.  THEOR. 

If  two  similar  parallelograms  have  a  common  angle, 
and  be  similarly  situated,  they  are  about  the  same  di- 
ameter. 

Let  the  parallelograms  ABCD,  AEFG  be  similar  and  similarly- 
situated,  and  have  the  angle  DAB  common:  ABCD  and  AEFG 
are  about  the  same  diameter. 

For,  if  not,  let,  if  possible,  the  paral-         AG  D 

lelogram  BD  have  its  diameter  AHC  in 
a  different  straight  line  from  AF  the  jr 
diameter  of  the  parallelogram  EG,  and 
let  GF  meet  AHC  in  H;  and  through  E 
H  draw  HK  parallel  to  AD  or  BCj 
therefore  the  parallelograms  ABCD, 
AKHG  being  about  the  same  diameter, 
they  are  similar  to  one  another  (24.  6.): 
wherefore  as  DA  to  AB,  so  is  (l.  def.         B  C 

6.)  GA  to  AK:  but  because  ABCD  and  AEFG  are  similar  paral- 
lelograms, as  DA  is  to  AB,  so  is  GA  to  AE;  therefore  (11.  5.)  a^ 
GA  to  AE,  so  GA  to  AK;  wherefore  GA  has  the  same  ratio  to 
each  of  the  straight  lines  AE,  AK;  and  consequently  AK  is  equal 
(9.  5.)  to  AE,  the  less  to  the  greater,  which  is  impossible:  there- 
fore ABCD  and  AKHG  are  not  about  the  same  diameter;  where- 
fore ABCD  and  AEFG  must  be  about  the  same  diameter.  There- 
fore, if  two  similar,  8cc.  Q.  E.  D. 

'  To  understand  the  three  following  propositions  more  easily, 
it  is  to  be  observed. 

'  1.  That  a  parallelogram  is  said  to  be  applied  to  a  straight 
line,  when  it  is  described  upon  it  as  one  of  its  sides.  Ex.  gr.  the 
parallelogram  AC  is  said  to  be  applied  to  the  straight  line  AB. 

*  2.  But  a  parallelogram  AE  is  said  to  be  applied  to  a  straight 
line  AB,  deficient  by  a  parallelogram,  when  AD  the  base  of  AE 


XH 

^ 

\ 

F 

E     C 


G 


D     B 


is  less  than  AB,  and  therefore  AE  is 
less  than  the  parallelogram  AC  de- 
scribed upon  AB  in  the  same  angle,  and 
between  the  same  parallels,  by  the  pa- 
rallelogram DC;  and  DC  is  therefore 
called  the  defect  of  AE. 

*  3.  And  a  parallelogram  AG  is  said  to  be  applied  to  a  straight 
line  AB,  exceeding  by  a  parallelogram,  when  AF  the  base  of  AG 
is  greater  than  AB,  and  therefore  AG  exceeds  AC  the  parallelo- 
gram described  upon  AB  in  the  same  angle,  and  between  the  same 
parallels,  by  the  same  parallelogram  BG.' 


148 


THE  ELEMENTS  OF  EUCLID. 


BOOK  VI. 


PROP.  XXVII.  THEOR. 

Of  all  parallelograms  applied  to  the  same  straight 
line,  and  deficient  by  parallelograms,  similar  and  simi- 
larly situated  to  that  which  is  described  upon  the  half 
of  the  hne;  that  which  is  applied  to  the  half  and  is  simi- 
lar to  its  defect,  is  the  greatest.* 

Let  AB  be  a  straight  line  divided  into  equal  parts  in  C; 
and  let  the  parallelogram  AD  be  applied  to  the  half  AC,  which 
is  therefore  deficient  from  the  parallelogram  upon  the  whole  line 
AB  by  the  parallelogram  CE  upon  the  other  half  CB:  of  all  the 
parallelograms  applied  to  any  other  parts  of  AB,  and  deficient  by 
parallelograms  that  are  similar  and  similarly  situated  to  CE:  AD 
is  the  greatest. 

Let  AF  be  any  parallelogram  applied  to  AK,  any  other  part  of 
AB  than  the  half,  so  as  to  be  deficient  from  the  parallelogram 
upon  the  whole  line  AB  by  the  parallelogram  KH  similar,  and  si- 
milarly situated  to  CE.    AD  is  greater  than  AF. 

First,  let  AK  the  base  of  AF,  be  greater  than  AC  the  half  of 
AB;  and  because  CE  is  similar  to  the 
parallelogram  KH,  they  are  about  the 
same  diameter  (26.  6.):  draw  their  dia- 
meter DB,  and  complete  the  scheme: 
because  the  parallelogram  CF  is  equal 
(43.  1.)  to  FE,  add  KH  to  both,  there- 
fore the  whole  CH  is  equal  to  the  whole 
KE:  but  CH  is  equal  (36.  1.)  to  CG, 
because  the  base  AC  is  equal  to  the 
baseCB:  therefore  CG  is  equal  to  KE:         A  C    K 

to  each  of  these  add  CF;  then  the  whole  AF  is  equal  to  the  gno- 
mon CHL:  therefore  CE,  or  the  parallelogram  AD,  is  greater 
than  the  parallelogram  AF. 

Next,  let  AK  the  base  of  AF  be  less 
than  AC,  and,  the  same  construction 
being  made,  the  parallelogram  DH  is 
equal  to  DG  (36.  1.),  for  HM  is  equal 
to  MG  (34.  1.)  because  BC  is  equal  to 
CA;  wherefore  DH  is  greater  than  LG: 
but  DH  is  equal  (43.  1.)  to  DK;  there- 
fore DK  is  greater  than  LG;  to  each 
of  these  add  AL;  then  the  whole  AD 
is  greater  than  the  whole  AF.  There- 
fore of  all  parallelograms  applied,  &c. 
Q.  E.  D. 


G 


G      F    M 


H 


*•  See  Note. 


BOOK    VI.  THE    ELEMENTS    OF    EUCLID.  149 

PROP.  XXVIII.  PROB. 

To  a  given  straight  line  to  apply  a  parallelogram 
equal  to  a  given  rectilineal  figure,  and  deficient  by  a 
parallelogram  similar  to  a  given  parallelogram:  but  the 
given  rectilineal  figure  to  which  the  parallelogram  to 
be  apphed  is  to  be  equal,  must  not  be  greater  than  the 
parallelogram  applied  to  half  of  the  given  line,  having 
its  defect  similar  to  the  defect  of  that  which  is  to  be 
applied :  that  is,  to  the  given  parallelogram.^ 

Let  AB  bQ,lhe  given  straight  line,  and  C  the  given  rectilineal 
figure,  to  which  the  parallelogram  to  be  applied  is  required  to 
be  equal,  which  figure  must  not  be  greater  than  the  parallelo- 
gram applied  to  the  half  of  the  line  having  its  defect  from  that 
upon  the  whole  line  similar  to  the  defect  of  that  which  is  to  be 
applied;  and  let  D  be  the  parallelogram  to  which  this  defect  is 
required  to  be  similar.  It  is  required  to  apply  a  parallelogram 
to  the  straight  line  AB,  which  tt  G      O    F 

shall  be  equal  to  the  figure 
C,  and  be  deficient  from  the 
parallelogram  upon  the  whole 
line  by  a  parallelogram  simi- 
lar to  D. 

Divide  AB  in  two  equal 
parts  (10.  1.)  in  the  point  E, 
and  upon  EB  describe  the 
parallelogram  EBFG  similar 
(18.  6.)  and  similarly  situated 
to  D,  and  complete  the  pa- 
rallelogram AG,  which  must 
either  be  equal  to  C  or  greater 
than  it,  by  the  determination:  '  ^      ^ 

and  if  AG  be  equal  to  C,  then  what  was  required  is  already  done: 
for,  upon  the  straight  line  AB,  the  parallelogram  AG  is  applied 
equal  to  the  figure  C,  and  deficient  by  the  parallelogram  EF  si- 
milar to  D:  but,  if  AG  be  not  equal  to  C,  it  is  greater  than  it: 
and  EF  is  equal  to  AG;  therefore  EF  also  is  greater  than  C. 
Make  (25.  6.)  the  parallelogram  KLMN  equal  to  the  excess  of  EF 
above  C,  and  similar  and  similarly  situated  to  D;  but  D  is  simi- 
lar to  EF,  therefore  (21.  6.)  also  KM  is  similar  to  EF;  let  KL 
be  the  homologous  side  to  EG,  and  LM,  to  GF:  and  because  EF 
is  equal  to  C  and  KM  together,  EF  is  greater  than  KM;  therefore 
the  straight  line  EG  is  greater  than  KL,  and  GF  than  LM:  make 
GX  equal  to  LK,  and  GO  equal  to  LM,  and  complete  the  paral- 
lelogram XGOP:  therefore  XO  is  equal  and  similar  to  KM;  but 
KM  is  similar  to  EF;  wherefore  also  XO  is  similar  to  EF,  and 
therefore  XO  and  EF  are  about  the  same  diameter  (26.  6.):  let 
GPB  be  their  diameter,  and  complete  the  scheme:  then  because 

*  See  Note. 


150 


THE  ELEMENTS  OF  EUCLID. 


BOOK  VI. 


EF  is  equal  to  C  and  KM  together,  and  XO  a  part  of  the  one  is 
equal  to  KM  a  part  of  the  other,  the  remainder,  viz.  the  gnomon 
ERO,  is  equal  to  the  remainder  C:  and  because  OR  is  equal  (34. 
I.)  to  XS,  Ijy  adding  SR  to  each,  the  whole  OB  is  equal  to  the 
whole  XB:  but  XB  is  equal  (36.  1.)  to  TE,  because  the  base  AE 
is  equal  to  the  base  EB,-  wherefore  also  TE  is  equal  to  OB:  add 
XS  to  each,  then  the  whole  TS  is  equal  to  the  whole,  viz.  to  the 
gnomon  ERO:  but  it  has  been  proved  that  the  gnomon  ERO  is 
equal  to  C,  and  therefore  also  TS  is  equal  to  C.  Wherefore  the 
parallelogram  TS,  equal  to  the  given  rectilineal  figure  C,  is  ap- 
plied to  the  given  straight  line  AB  deficient  by  the  parallelogram 
SR,  similar  to  the  given  one  D,  because  SR  is  similar  to  EF  (24. 
6.).     Which  was  to  be  done. 

PROP.  XXIX.  PROB. 

To  a  given  straight  line  to  apply  a  parallelogram 
equal  to  a  given  rectilineal  figure,  exceeding  by  a  paral- 
lelogram similar  to  another  given.^ 

Let  AB  be  the  given  straight  line,  and  C  the  given  rectilineal 
figure  to  which  the  parallelogram  to  be  applied  is  required  to  be 
equal,  and  D  the  parallelogram  to  which  the  excess  of  the  one  to 
be  applied  above  that  upon  the  given  line  is  required  to  be  simi- 
lar. It  is  required  to  apply  a  parallelogram  to  the  given  straight 
line  AB,  which  shall  be  equal  to  the  figure  C,  exceeding  by  a  pa- 
rallelogram similar  to  D. 

Divide  AB  into  two  equal  parts  in  the  point  E,  and  upon  EB 
describe  (18.  6.)  the  parallelogram  EL  similar  and  similarly  si- 
tuated to  D:  and  make  (25.  6.)  the  parallelogram  GH  equal  to  EL 
and  C  together,  and  similar  and  similarly  situated  to  D;  where- 
fore GH  is  similar  to  EL  (21.  6.);  let  KH  be  the  side  homologous 
to  FL,  and  KG  to  FE;  and  because  the  parallelogram  GH  is 
greater  than  EL,  therefore  the  side  KH  is  greater  than  FL,  and 
KG  than  FE:  produce  FL  and  FE,  and  make  FLM  equal  to  KH, 
and  FEN  to  KG,  and  complete  the  parallelogram  MN.  MN  is 
therefore  equal  and  si- 
milar to  GH,  but  GH 
is  similar  to  EL; 
wherefore  MN  is  si- 
milar to  EL,  and  con- 
sequently EL  and  MN 
are  about  the  same  di- 
ameter (26.  6.):  draw 
their  diameter  FX, 
and  complete  the 
scheme.  Therefore, 
since  GH  is  equal  to 
EL  and  C  together, 
and  that  GH  is  equal 
to  MN;  MN  is  equal 


See  Note. 


BOOK  VI.  THE  ELEMENTS  OF  EUCLID.  I5i 

to  EL  and  C:  take  away  the  common  part  EL:  then  the  remain- 
der, viz.  the  gnomon  NOL,  is  equal  to  C.  And  because  AE  is 
equal  to  EB,  the  parallelogram  AN  is  equal  (36.  1.)  to  the  paral- 
lelogram NB,  that  is,  to  BM  (43.  1.).  Add  NO  to  each;  there- 
fore the  whole,  viz.  the  parallelogram  AX  is  equal  to  the  gnomon 
NOL.  But  the  gnomon  NOL  is  equal  to  C;  therefore  also  AX 
is  equal  to  C.  Wherefore  to  the  straight  line  AB  there  is  applied 
the  parallelogram  AX  equal  to  the  given  rectilineal  C,  exceeding 
by  the  parallelogram  PO,  which  is  similar  to  D,  because  PO  is 
similar  to  EL  (24.  6.)     Which  Was  to  be  done. 

PROP.  XXX.  PROB. 

To  cut  a  given  straight  line  in  extreme  and  mean 
ratio. 

Let  AB  be  jf^  given  straight  line;  it  is  required  to  cut  it  in 
extreme  and  mean  ratio. 

Upon  A^ 'describe  (46.  1.)  the  square  BC,  and  to  AC  apply 
the  parallelogram  CD  equal  to  BC,  exceeding  by  the  figure  AD 
similar  to  BC  (29.  6.):  but  BC  is  a  square, 
therefore  also  AD  is  a  scpiare;  and  because 
BC  is  equal  to  CD,  by  taking  the  common 

part  CE   from  each,   the  remainder   BF  is   A  ] (^         B 

equal  to  the  remainder  AD:  and  these 
figures  are  equiangular,  therefore  their  sides 
about  the  equal  angles  are  reciprocally  pro- 
portional (14.6.);  wherefore  as  FE  to  ED, 
so  AE  to  EB;  but  FE  is  equal  to  AC  (34.  1.), 
that  is,  to  AB;  and  ED  is  equal  to  AE:  there- 
fore as  BA  to  AE,  so  is  AE  to  EB:  but  AB 
is  greater  than  AE;  wherefore  AE  is  greater 

than  EB  (14.  5.):  therefore  the  straight  line  AB  is  cut  in  extreme 
and  mean  ratio  in  E  (3.  def.  6.).     Which  was  to  be  done. 

Otherwise. 

Let  AB  be  the  given  straight  line;  it  is  required  to  cut  it  in  ex^" 
treme  and  mean  ratio. 

Divide  AB  in  the  point  C,  so  that  the  rectangle  contained  by 

AB,  BC  be  equal  to  the  square  of  AC  (U.  2.). 1 

Then,  because  the  rectangle  AB,  BC  is  equal  to     A  C     B 

the  square  of  AC,  as  BA  to  AC,  so  is  AC  to  CB  (17.  6.):  there- 
fore AB  is  cut  in  extreme  and  mean  ratio  in  C  (3.  def.  6.).  Which 
was  to  be  done. 

PROP.  XXXL  THEOR. 

In  right  angled  triangles,  the  rectilineal  figure  de- 
scribed upon  the  side  opposite  to  the  right  angle,  is 
equal  to  the  similar  and  similarly  described  figures  upon 
the  sides  containing  the  right  angle.* 

""  See  Note. 


e;- 

— ^                      .■- 

152 


THE  ELEMENTS  OF  EUCLID. 


BOOK  VI.. 


Let  ABC  be  a  right  angled  triangle,  having  the  -righti angle 
BAG;  the  rectilineal  figure  described  upon  BC  is  equal  to  the 
similar  and  similarly  described  figures  upon  BA,  AC. 

Draw  the  perpendicular  AD^  therefore,  because  in  the  right 
angled  triangle  ABC,  AD  is  drawn  from  the  right  angle  at  A 
perpendicular  to  the  base  BC,  the  triangles  ABD,  ADC  are  si- 
milar to  the  whole  triangle  ABC,  and  to  one  another  (8.  6.),  and 
because  the  triangle  ABC  is  similar  to  ADB,  as  CB  to  BA,  so 
is  BA  to  BD  (4.  6.);  and  because  these  three  straight  lines  are 
proportionals,  as  the  first  to  the  third,  so  is  the  figure  upon  the 
first  to  the  similar  and  similarly  described  figure  upon  the  se- 
cond (2  Cor.):  therefore  as  CB  to 
BD,  so  is  the  figure  upon  CD  to 
the  similar  and  similarly  described 
figure  upon  BA:  and,  inversely  (B. 
5.),  as  DB  to  BC,  so  is  the  figure 
upon  BA  to  that  upon  BC;  for  the 
same  reason,  as  DC  to  CB,  so  is 
the  figure  upon  CA  to  that  upon 

CB.     Wherefore   as  BD  and   DC 

together  to  BC,  so  are  the  figures  upon  BA,  AC  to  that  upon  BC 
(24.  5.):  but  BD  and  DC  together  are  equal  to  BC.  Therefore 
the  figure  described  on  BC  is  equal  (A.  5.)  to  the  similar  and  si- 
milarly described  figures  on  BA,  AC.  Wherefore,  in  right  angled 
triangles,  Sec.     Q.  E.  D. 


PROP.  XXXII.  THEOR. 

If  two  triangles  which  have  two  sides  of  the  one  pro- 
portional to  two  sides  of  the  other,  be  joined  at  one  an- 
gle, so  as  to  have  their  homologous  sides  parallel  to  one 
another,  the  remaining  sides  shall  be  in  a  straight  line.^ 

Let  ABC,  DCE  be  two  triangles  which  have  the  two  sides 
BA,  AC  proportional  to  the  two  CD,  DE,  viz.  BA  to  AC,  as 
CD  to  DE;  and  let  AB  be  parallel  to  DC,  and  AC  to  DE.  BC 
and  CE  are  in  a  straight  line. 

Because  AB  is  parallel  to  DC,  A' 
and  the  straight  line  AC  meets 
them,  the  alternate  angles  BAC, 
ACD  are  equal  (29.  I.);  for  the 
same  reason,  the  angle  CDE  is 
equal  to  the  angle  ACD;  where- 
fore also  BAC  is  equal  to  CDE: 
and  because  the  triangles  ABC,  B  C  E 

DCE  have  one  angle  at  A  equal  to  one  at  D,  and  the  sides  about 
these  angles  proportionals,  viz.  BA  to  AC,  as  CD  to  DE,  the 
triangle  ABC  is  equiangular  (6.  6.)  to  DCE:  therefore  the  angle 
ABC  is  equal  to  the  angle  DCE;  and  the  angle  BAC  was  proved 
to  be  equal  to  ACD:  therefore  the  whole  angle  ACE  is  equal  to 


*  « 


See  Note. 


■'■i^^:.. 


BOOK  vr. 


THE  ELEMENTS  OF  EUCLID. 


153 


the  two  angles  ABC,  BAG 5  add  the  common  angle  ACB,  then 
the  atigles  ACE,  ACB  are  equal  to  the  angles  ABC,  BAC,  ACB: 
but  ABC,  BAC,  ACB  are  equal  to  two  right  angles  (32.  l.)^ 
therefore  also  the  angles  ACE,  ACB  are  equal  to  two  right  an- 
gles: and  since  at  the  point  C,  in  the  straight  line  AC,  the  two 
straight  lines  BC,  CE,  which  are  on  the  opposite  sides  of  it,  make 
the  adjacent  angles  ACE,  ACB  equal  to  two  right  angles;  there- 
fore (14.  1.)  BC  and  CE  are  in  a  straight  line.  Wherefore,  if  two 
triangles,  &c.  Q.  E.  D. 

PROP.  XXXIII.  THEOR. 

In  equal  circles,  angles,  whether  at  the  centres  or  cir- 
cumferences, have  the  same  ratio  which  the  circumfer- 
ences on  which  they  stand  have  to  one  another:  so 
also  have  the  sectors.* 

Let  ABC,  DEF  be  equal  circles:  and  at  their  centres  the  angles 
BGC,  EHF,  and  the  angles  BAC,  EDF  at  their  circumferences: 
as  the  circumference  BC  to  the  circumference  EF,  so  is  the  angle 
BGC  to  the  angle  EHF,  and  the  angle  BAC  to  the  angle  EDF; 
and  also  the  sector  BGC  to  the  sector  EHF. 

Take  any  number  of  circumferences  CK,  KL,  each  equal  to  BC, 
and  any  number  whatever  FM,  MN,  each  equal  to  EF:  and  join 
GK,  GL,  HM,  HN.  Because  the  circumferences  BC,  CK,  KL 
are  all  equal,  the  angles  BGC,  CGK,  KGL  are  also  all  equal  (27. 
3.):  therefore  what  multiple  soever  the  circumference  BL  is  of 
the  circumference  BC,  the  same  multiple  is  the  angle  BGL  of  the 
angle  BGC:  for  the  same  reason,  whatever  multiple  of  the  cir- 
cumference EN  is  of  the  circumference  EF,  the  same  multiple  is 
the  angle  EHN  of  the  angle  EHF:  and  if  the  circumference  BL 
be  equal  to  the  circumference  EN,  the  angle  BGL  is  also  equal 
(27.  3.)  to  the  angle  EHN;  and  if  the  circumference  BL  be 
greater  than  EN,  likewise  the  angle  BGL  is  greater  than  EHN; 
and  if  less,  less:  there  being  then  four  magnitudes,  the  two  cir- 
cumferences BC,  EF,  and  the  two  angles  BGC,  EHF;  of  the  cir- 
cumference BC,  and  of  the  angle  BGC,  have  been  taken  any  equi- 
multiples whatever,  viz.  the  circumference  BL,  and  the  angle  BGL; 
and  of  the  circumference  EF,  and  of  the  angle  EHF,  any  equimul- 


'  See  Note. 


154  THE  ELEMENTS  OF  EUCLID.  BOOK   VI. 

tiples  whatever,  viz.  the  circumference  EN,  and  the  Single  EHN: 
and  it  has  been  proved,  that  if  the  circumference  BL  be  greater' 
than  EN,  the  angle  BGL  is  greater  than  EHN:  and  if  equal, 
equals  and  if  less,  less:  as  therefore  the  circumference  BC  to  the 
circumference  EF,  so  (5.  def.  5.)  is  the  angle  BGC  to  the  angle 
EHF:  but  as  the- angle  BGC  is  to  ihe  angle  EHF,  so  is  (15.  5.) 
the  angle  BAG  to  the  angle  EDF,  for  each  is  double  of  each  (20„ 
3.):  therefore,  as  the  circumference  BC  is  to  EF,  so  is  the  angle 
BGC  to  the  angle  EHF,  and  the  angle  BAG  to  the  angle  EDF. 

Also,  as  the  circumference  BC  to  EF,  so  is  the  sector  BGC  to 
the  sector  EHF.     Join  BC,  CK,  and  in  the  circumferences  BC, 
CK  take  any  points  X,  O,  and  join  BX,  XC,  CO,  OK:  then,  be- 
cause in  the  triangles  GBC,  GCK,  the  two  sides  BG,  GC  are 
equal  to  the  two  CG,  GK,  and  that  they  contain  equal  angles^ 
the  base  BC  is  equal  (4.  1.)  to  the  base  CK,  and  the  triangle  GBC 
to  the  triangle  GCK:  and  because  the  circumference  BC  is  equal 
to  the  circumference  CK,  the  remaining  part  of  the  whole  circum- 
ference of  the  circle  ABC,  is  equal  to  the  remaining  part  of  the 
whole  circumference  of  the  same  circle:  wherefore  the  angle  BXC 
is  equal  to  the  angle  COK  (27.  3.),  and  the  segment  BXC  is  there- 
fore similar  to  the  segment  COK  (11.  def.  3.):  and  they  are  upon 
equal  straight  lines  BC,  CK:  but  similar  segments  of  circles  upon 
equal  straight  lines,  arc  equal  (24.  3.)  to  one  another:  therefore 
the  segment  BXC  is  equal  to  the  segment  COK:  and  the  triangle 
BGC  is  equal  to  the  triangle  CGK;  therefore  the  whole,  the  sec- 
tor BGC,  is  equal  to  the  whole,  the  sector  CGK:  for  the  same  rea- 
son, the  sector  KGL  is  equal  to  each  of  the  sectors  BGC,  CGK: 
in  the  same  manner,  the  sectors  EHF,  FHM,  MHN  may  be  proved 
equal  to  one  another:  therefore,  what  multiple  soever  the  circum- 
ference BL  is  of  the  circumference  BC,  the  same  multiple  is  the 
sector  BGL  of  the  sector  BGC:  for  the  same  reason,  whatever 
multiple  the  circumference  EN  is  of  EF,  the  same  multiple  is  the 
sector  EHN  of  the  sector  EHF:  and  if  the  circumference  BL  be 
equal  to  EN,  the  sector  BGL  is  equal  to  the  sector  EHN:  and  if 


the  circumference  BL  be  greater  than  EN,  the  sector  BGL  is 
greater  than  the  sector  EHN;  and  if  less,  less:  since,  then,  there 
are  four  magnitudes,  the  two  circumferences  BC,  EF,  and  the  two 
sectors  BGC,  EHF,  and  of  the  circumference  BC,  and  sector 
BGC,  the  circumference  BL  and  sector  BGL  are  any  equal  mul- 


BOOK  VI.  THE  ELEMENTS  OF  EUCLID.  155 

tiples  whatei'er:  and  of  the  circumference  EF,  and  sector  EHF, 
the  circumference  EN  and  sector  EHN  are  any  equimultiples 
whatever^  and  that  it  has  been  proved,  if  the  circumference  BL 
be  greater  than  EN,  the  sector  BGL  is  greater  than  the  sector 
EHN^  and  if  equal,  equals  and  if  less,  less.  Therefore  (5.  def. 
5.),  as  the  circumference  BC  is  to  the  circumference  EF,  so  is  the 
sector  BGC  to  the  sector  EHF.  Wherefore,  in  equal  circles,  8cc. 
Q.  E.  D. 

PROP.  B.  THEOR. 

If  an  angle  of  a  triangle  be  bisected  by  a  straight 
line,  which  likewise  cuts  the  base;  the  rectangle  con- 
tained by  the  sides  of  the  triangle  is  equal  to  the  rect- 
angle contained  by  the  segments  of  the  base,  together 
with  the  square  of  the  straight  line  bisecting  the  angle.* 

Let  ABC  be  a  triangle,  and  let  the  angle  BAG  be  bisected  by 
the  straight  line  AD;  the  rectangle  BA,  AC  is  equal  to  the  rect- 
angle BD,  DC  together  with  the  square  of  AD. 

Describe  the  circle  (5.  4.)  ACB  about  the  triangle,  and  pro- 
duce AD  to  the  circumference  in  E, 
and  join  EC:  then  because  the  aTig;lc 
BAD  is  equal  to  the  angle  CAE,  and 
the  angle  ABD  to  the  angle  (21.  3.) 
AEC,  for  they  arc  in  the  same  seg-  ^ 
ment:  the  triangles  ABD,  AEC,  are 
equiangular  to  one  another:  therefore 
as  BA  to  AD,  so  is  (4.  6.)  EA  to  AC, 
and  consequently  the   rectangle  BA, 

AC  is  equal  (16.  6.)  to  the  rectangle  

EA,  AD,  that  is  (3.  2.),  to  the  rectan-  E 

gle  ED,  DA,  together  with  the  square  of  AD:  but  the  rectangle 
ED,  DA  is  equal  to  the  rectangle  (35.  3.)  BD,  DC.  Therefore 
the  rectangle  BA,  AC  is  equal  to  the  rectangle  BD,  DC,  together 
with  the  square  of  AD.     Wherefore,  if  an  angle.  Sec.  Q.  E.  D. 

PROP.  C.  THEOR. 

If  from  any  angle  of  a  triangle  a  straight  line  be 
drawn  perpendicular  to  the  base;  the  rectangle  con- 
tained by  the  sides  of  the  triangle  is  equal  to  the  rect- 
angle contained  by  the  perpendicular  and  the  diameter 
of  the  circle  described  about  the  triangle.* 

Let  ABC  be  a  triangle,  and  AD  the  perpendicular  from  the 
angle  A  to  the  base  BC;  the  rectangle  BA,  AC  is  equal  to  the 
rectangle  contained  by  AD  and  the  diameter  of  the  circle  de- 


scribed about  the  triangle. 


Sec  Note. 


156 


THE  ELEMENTS  OF  EUCLID. 


BOOK  VI. 


Describe  (5.  4.)  the  circle  ACB 
about  the  triangle,  and  draw  its  dia- 
meter AE,  and  join  EC;  because  the 
right  angle  BDA  is  equal  (31.  3.)  to 
the  angle  EGA  in  a  semicircle,  and 
the  angle  ABD  to  the  angle  AEC  in 
the  same  segment  (21.  3.);  the  tri- 
angles ABD,  AEC  are  equiangular: 
therefore  as  (4.  6.)  BA  to  AD,  so  is 
EA  to  AC;  and  consequently  the  rect- 
angle BA,  AC  is  equal  (16.  6.)  to  the 
rectangle  EA,  AD.     If,  therefore,  from  an  angle,  &c.  Q.  E.  D. 


PROP.  D.  THEOR. 

The  rectangle  contained  by  the  diagonals  of  a  quad- 
rilateral inscribed  in  a  circle,  is  equal  to  both  the  rect- 
angles contained  by  its  opposite  sides.* 

Let  ABCD  be  any  quadrilateral  inscribed  in  a  circle,  and  join 
AC,  BDj  the  rectangle  contained  by  AC,  BD  is  equal  to  the  two 
rectangles  contained  by  AB,  CD,  and  by  AD,  BC.f 

Make  the  angle  ABE  equal  to  the  angle  DBC;  add  to  each 
of  these  the  common  angle  EBD,  then  the  angle  ABD  is  equal 
to  the  angle  EBCj  and  the  angle  BDA  is  equal  (21.  3.)  to  the 
angle  BCE,  because  they  are  in  the  same  segment;  therefore  the 
triangle  ABD  is  equiangular  to  the  g 

triangle  BCE;  wherefore  (4.  6.)  as 

BC  is  to  CE,  so  is  BD  to  DA;  and  /^;- ___    \  C 

consequently  the  rectangle  BC,  AD 

is  equal  (16. 6.)  to  the  rectangle  BD, 

CE:  again,  because  the  angle  ABE 

is  equal  to  the  angle  DBC,  and  the 

angle  (2 1 .  3.)  B  AE  to  the  angle  BDC, 

the  triangle  ABE  is  equiangular  to 

the  triangle  BCD:  as  therefore  BA 

to  AE,  so  is  BD  to  DC;  wherefore 

the  rectangle  BA,  DC  is  equal  to  the 

rectangle  BD,  AE:  but  the  rectangle  BC,  AD  has  been  shown 

equal  to  the  rectangle  BD,  CE;  therefore  the  whole  rectangle  AC, 

BD  (1.  2.)  is  equal  to  the  rectangle  AB,  DC,  together  with  the 

rectangle  AD,  BC.    Therefore  the  rectangle,  &c.  Q.  E.  D. 


*  See  Note. 

1  This  is  a  Lemma  of  CI.  Ptoloraseus,  in  page  9  of  his  jutyAhn  o-vvrA^tc- 


THE 


EliEMESfTS  OF  EUCJLID. 


BOOK  XI. 


DEFINITIONS. 


I. 

A  SOLID  is  that  which  hath  length,  breadth,  and  thickness. 

II. 

That  which  bounds  a  solid  is  a  superficies. 

III. 

A  straight  line  is  perpendicular,  or  at  right  angles  to  a  plane, 
when  it  makes  right  angles  with  every  straight  line  meeting  it 
in  that  plane. 

IV. 

A  plane  is  perpendicular  to  a  plane,  when  the  straight  lines 
drawn  in  one  of  the  planes  perpendicularly  to  the  common  sec- 
tion of  the  two  planes,  are  perpendicular  to  the  other  plane. 

V. 

The  inclination  of  a  straight  line  to  a  plane  is  the  acute  angle  con- 
tained by  that  straight  line,  and  another  drawn  from  the  point 
in  which  the  first  line  meets  the  plane  to  the  point  in  which  a 
perpendicular  to  the  plane  drawn  from  any  point  of  the  first  line 
above  the  plane,  meets  the  same  plane. 

VI. 

The  inclination  of  a  plane  to  a  plane  is  the  acute  angle  contained 
by  two  straight  lines  drawn  from  any  the  same  point  of  their 
common  section  at  right  angles  to  it,  one  upon  one  plane,  and 
the  other  upon  the  other  plane. 

VII. 

Two  planes  are  said  to  have  the  same,  or  a  like  inclination  to  one 
another,  which  two  other  planes  have,  when  the  said  angles  of 
inclination  are  equal  to  one  another. 


158  THE  ELEMENTS  OF  EUCLID.  BOOK  XI. 

VIII. 

Parallel  planes  are  such  which  do  not  meet  one  another  though 

produced. 

IX. 
A  solid  angle  is  that  which  is  made  by  the  meeting  of  more  than 

two  plane  angles,  which  are  not  in  the  same  plane,  in  one  point.* 

X. 

*The  tenth  definition  is  omitted  for  reasons  given  in  the  notes.* 

XI. 

Similar  solid  figures  are  such  as  have  all  their  solid  angles  ^qual, 
each  to  each,  and  which  are  contained  by  the  same  number  of 
similar  planes.* 

XII. 
A'  pyramid  is  a  solid  figure  contained  by  planes  that  are  consti- 
tuted betwixt  one  plane  and  one  point  above  it  in  which  they 
meet. 

XIII. 
*■  A  prism  is  a  solid  figure  coift^ained  by  plane  figures,  of  which 
two   that  are  opposite  are  equal,  similar,  and  parallel  to  one 
another^  and  the  others  parallelograms. 

XIV. 

A  sphere  is  a  solid  figure  described  by  the  revolution  of  a  semi- 
circle about  its  diameter,  which  remains  unmoved. 

XV. 
The  axis  of  a  sphere  is  the  fixed  straight  line  about  which  the 
semicircle  revolves. 

XVI. 
The  centre  of  a  sphere  is  the  same  with  that  of  a  semicircle. 

XVII. 

The  diameter  of  a  sphere  is  any  straight  line  which  passes 
through  the  centre,  and  is  terminated  both  ways  by  the  super^ 
ficies  of  the  sphere. 

XVIII. 

A  cone  is  a  solid  figure  described  by  the  revolution  of  a  right 
angled  triangle  about  one  of  the  sides  containing  the  right  an- 
gle, which  side  remains  fixed. 

If  the  fixed  side  be  equal  to  the  other  side  containing  the  right 
angle,  the  cone  is  called  a  right  angled  cone;  if  it  be  less  than 
the  other  side,  an  obtuse  angled,  and  if  greater,  an  acute  angled 
cone. 

XIX. 

The  axis  of  a  cone  is  the  fixed  straight  line  about  which  the  trian- 
gle revolves. 

XX. 

The  base  of  a  cone  is  the  circle  described  by  that  side  containing 
the  right  angle,  which  revolves. 

XXI. 

A  cylinder  is  a  solid  figure  described  by  the  revolution  of  a  right 
angled  parallelogram  about  one  of  its  sides  which  remains 
fixed. 

^  See  Note. 


BOOK  XI.  THE  ELEMENTS  OF  EUCLID.  159 

XXII. 

The  axis  of  a  cylinder  is  the  fixed  straight  line  about  which  the 
parallelogram  revolves. 

XXIII. 

The  bases  of  a  cylinder  are  the  circles  described  by  the  two  re- 
volving opposite  sides  of  the  parallelogram. 

XXIV. 

Similar  cones  and  cylinders  are  those  which  have  their  axes  and 
the  diameters  of  their  bases  proportionals. 

XXV. 

A  cube  is  a  solid  figure  contained  by  six  equal  squares. 

XXVI. 

A  tetrahedron  is  a  solid  figure  contained  by  four  equal  and  equi- 
lateral triangles. 

XXVII. 
An  octahedron  is  a  solid  figure  contained   by  eight  equal   and 
equilateral  triangles.  -  -* 

XXVIII. 
A  dodecahedron  is  a  solid  figure  contained  by  twelve  equal  pen- 
tagons which  are  equilateral  and  equiangular. 

^  XXIX. 

An  icosahedron  is  a  solid  figure  contained  by  twenty  equal  and 

equilateral  triangles. 

DEF.  A. 
A  parallelopiped  is  a  solid  figure  contained  by  six  quadrilateral 

figures,  whereof  every  opposite  two  are  parallel. 

PROP.  I.  THEOR. 

One  part  of  a  straight  line  cannot  be  in  a  plane,  and 
another  part  above  it.* 

If  it  be  possible,  let  AB,  part  of  the  straight  line  ABC,  be  in 
the  plane,  and  the  part  BC  above  it:  and  since  the  straight  line 
AB  is  in  the  plane,  it  can  be  pro- 


duced in  that  plane:  let  it  be  pro-     <r 

duced  to  D:  and  let  any  plane  pass      \ 

through  the  straight  line  AD,  and        \  A J^ J^ 

be  turned  about  it  until  it  pass 
through  the  point  C:  and  because  the  points  B,  C  are  in  this 
plane,  the  straight  line  BC  is  in  it  (7.  def.  1.):  therefore  there  are 
two  straight  lines  ABC,  ABD  in  the  same  plane  that  have  a 
common  segment  AB,  which  is  impossibe  (Cor.  11.  1.).  There- 
fore, one  part,  &c.  Q.  E.  D. 

PROP.  II.  THEOR. 

Two  Straight  lines  which  cut  one  another  are  in  one 

*  See  Note. 


160 


THE  ELEMENTS  OF  EUCLID. 


BOOK  Xl.,fc 


plane,  and  three  straight  hnes  which  meet  one  another 
are  in  one  plane."^ 

Let  two  straight  lines,  AB,  CD  cut  one  another  in  E;  AB,  CD 
are  in  one  plane:  and  three  straight  lines  EC,  CB,  BE  which  meet 
one  another,  are  in  one  plane. 

Let  any  plane  pass  through  the  straight 
line  EB,  and  let  the  plane  be  turned  about 
EB,  produced,  if  necessary,  until  it  pass 
through  the  point  C:  then  because  the  points 
E,  C  are  in  this  plane,  the  straight  line  EC 
is  in  it  (7.  def.  1.):  for  the  same  reason,  the 
straight  line  BC  is  in  the  same;  and,  by  the 
hypothesis,  EB  is  in  it;  therefore  the  three 
straight  lines,  EC,  CB,  BE  are  in  one  plane; 
but  in  the  plane  in  which  EC,  EB  are,  in      C  B 

the  same  are  (l.  11.)  CD,  AB:  therefore  AB,  CD  are  in  one  plane. 
Wherefore  two  straight  lines,  Sec.  Q.  E.  D. 


PROP.  in.  THEOR. 

If  two  planes  cut  one  another,  their  common  section 
is  a  straight  line.^ 

Let  two  planes  AB,  BC  cut  one  another,  and  let  the  line  DB 
be  their  common  section:  DB  is  a  straight 
line:  if  it  be  not,  from  the  point  D  to  B, 
draw,  in  the  plane  AB,  the  straight  line 
DEB,  and  in  the  plane  BC  the  straight  line 
DFB:  then  two  straight  lines  DEB,  DFB 
have  the  same  extremities,  and  therefore  in- 
clude a  space  betwixt  them;  which  is  im- 
possible (10.  Ax.  1.):  therefore  BD  the  com- 
mon section  of  the  planes  AB,  BC  cannot 
but  be  a  straight  line.  Wherefore,  if  two 
planes,  &c.  Q.  E.  D. 

PROP.  IV.  THEOR. 

If  a  straight  line  stand  at  right  angles  to  each  of  two 
straight  lines  in  the  point  of  their  intersection,  it  shall 
also  be  at  right  angles  to  the  plane  which  passes 
through  them,  that  is,  to  the  plane  in  which  they  are.^ 

Let  the  straight  line  EF  stand  at  right  angles  to  each  of  the 
straight  lines  AB,  CD  in  E,  the  point  of  their  intersection:  EF  is 
also  at  right  angles  to  the  plane  passing  through  AB,  CD. 

Take  the  straight  lines  AE,  EB,  CE,  ED  all  equal  to  one 
another;  and  through  E  draw,  in  the  plane  in  which  are  AB, 
CD,  any  straight  line  GEH;  and  join  AD,  CB:  then,  from  any 
point  F  in  EF,  draw  FA,  FG,  FD,  FC,  FH,  FB:  and  because  the 


*  See  Note. 


BOOIC  XI. 


THE  ELEMENTS  OF  EUCLID. 


161 


two  straight  lilies,  AE,  ED  are  equal  to  the  two  BE,  EC,  and  that 
they  contain  equal  angles  (15.  1.)  AED,  BEC,  the  base  AD  is 
equal  (4.  1.)  to  the  base  BC,  and  the  angle  DAE  to  the  angle 
EEC:  and  the  angle  AEG  is  equal  to  the  angle  BEH  (15.  1.); 
therefore  the  triangles  AEG,  BEH  have  two  angles  of  one  equal  to 
two  angles  of  the  other,  each  to  each,  and  the  sides  AE,  EB,  ad- 
jacent to  the  equal  angles,  equal  to  one  another^  wherefore  they 
shall  have  their  other  sides  equal  (26.  1.):  GE  is  therefore  equal 
to  EHy  and  AG  to  BH:  and  because  AE  is  equal  to  EB,  and  FE 
common  and  at  right  angles  to  them,  the  base  AF  is  equal  (4.  1.) 
to  the  base  FB;  for  the  same  reason,  CF  is  equal  to  FD:  and  be- 
cause AD  is'equal  to  BC,  and  AF  to  FB,  the  two  sides  FA,  AD- 
are  equal  to  the  two  FB,  BC,  each  to  each,  ~ 

and  the  base  DP'  was  proved  equal  to  the 
base  EC^  tlierefore  the  angle  FAD  is  equal 
(87  1.)  to  the  angle  FBC:  again,  it  was 
proved  that  GA  is  equal  to  BH,  and  also 
AF  to  FB;  FA,  then,  and  AG  are  equal  to  A| 
F'B  and  BH,  and  the  angle  FAG  has  been 
proved  equal  to  the  angle  FBH;  there-  G 
fore  the  base  GF  (4.  1.)  to  the  base  FH: 
again,  because  it  was  proved,  that  GE  is 
equal  to  EH,  and  EF  is  common;  GE, 
EF  are  equal  to  HE,  EF;  and  the  base 
GF  is  equal  to  the  base  FFI;  therefore  the 
angle  GEF  is  equal  (8.  1.)  to  the  angle 
HEF;  and  consequently  each  of  these  angles  is  a  right  (10.  def.  1.) 
angle.  Therefore  FE  makes  right  angles  with  GH,  that  is,  with 
any  straight  line  drawn  through  E  in  the  plane  passing  through 
AB,  CD.  In  like  manner,  it  may  be  proved,  that  FE  makes  right 
angles  with  every  straight  line  which  meets  it  in  that  plane.  But 
a  straight  line  is  at  right  angles  to  a  plane  when  it  makes  right 
angles  with  every  straight  line  which  meets  it  in  that  plane  (3.  def. 
11.);  therefore  EF  is  at  right  angles  to  the  plane  in  which  are 
AB,  CD.     Wherefore,  if  a  straight  line,  &:c.     Q.  E.  D. 


PROP.  V.  THEOR. 

If  three  straight  hues  meet  all  in  one  point,  and  a 
straight  line  stands  at  right  angles  to  each  of  them  in 
that  point;  these  three  straight  lines  are  in  one  and  the 
same  plane."^ 

Let  the  straight  line  AB  stand  at  right  angles  to  each  of  the 
straight  lines  BC,  BD,  BE,  in  B  the  point  where  they  meet;  BC, 
BD,  BE  are  in  one  and  the  same  plane. 

If  not  let,  if  it  be  possible,  BD  and  BE  be  in  one  plane,  and 
BC  be  above  it;  and  let  a  plane  pass  through  AB,  BC,  the  com- 
mon section  of  which  with  the  plane,  in  vv'hich  BD  and  BE  are, 


*  See  Note, 


X 


162 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI. 


shall  be  a  straight  (3.  11.)  line;  let  this  be  BF:  therefore  the  three 
straight  lines  AB,  BC,  BF  are  all  in  one  plane,  viz.  that  which 
passes  through  AB,  BC:  and  because  AB  stands  at  right  angles 
to  each  of  the  straight  lines  BD,  BE,  it  is  also  at  right  angles 
(4.  11.)  to  the  plane  passing  through  them;  and  therefore  makes 
right  angles  (3.  def.  11.)  with  every 
straight  line  meeting  it  in  that  plane;  A 
but  BF,  which  is  in  that  plane,  meets 
it:  therefore  the  angle  ABF  is  a  right 
angle;  but  the  angle  ABC,  by  the  by-  .-.  /      F 

pothesis,  is  also  a  right  angle;  there- 
fore the  angle  ABF  is  equal  to  the  angle 
ABC,  and  they  are  both  in  the  same 
plane,  which  is  impossible:  therefore  the 
straight  line  BC  is  not  above  the  plane  in 
which  are  BD  and  BE:  wherefore  the  three  straight  lines  BC, 
BD,  BE  are  in  one  and  the  same  plane.  Therefore,  if  three 
straight  lines,  &cc.     Q.  E.  D. 


D 
E 


PROP.  VI.  THEOR. 

If  two  straight  lines  be  at  right  angles  to  the  same 
plane,  they  shall  be  parallel  to  one  another. 

Let  the  straight  lines  AB,  CD  be  at  right  angles  to  the  same 
plane;  AB  is  parallel  to  CD. 

Let  them  meet  the  plane  in  the  points  B,  D,  and  draw  the 
straight  line  B,D,to  which  draw  DE  at  right  angles,  in  the  same 
plane;  and  make  DE  equal  to  AB,  and  join 
BE,  AE,  AD.  Then  because  ^^AB  is  per- 
pendicular to  the  plane,  it  shall  make  right 
(3.  def.  11.)  angles  with  every  straight  line 
Avhich  meets  it,  and  is  in  that  plane:  but  BD, 
BE,  which  are  in  that  plane,  do  each  of  them 
meet  AB.  Therefore,  each  of  the  angles 
ABD,  ABE  is  a  right  angle:  for  the  same 
reason,  each  of  the  angles  CDB,  CDE  is  a 
right  angle:  and  because  AB  is  equal  to  DE, 
and  BD  common,  the  two  sides  AB,  BD  are 
equal  to  the  two  ED,  DB:  and  they  contain 
right  angles;  therefore  the  base  AD  is  equal  (4.  1.)  to  the  base 
BE:  again,  because  AB  is  equal  to  DE,  and  BE  to  AD;  AB,  BE 
are  equal  to  ED,  DA;  and,  in  the  triangles  ABE,  EDA,  the  base 
AE  is  common;  therefore  the  angle  ABE  is  equal  (8.  1.)  to  the 
angle  EDA:  but  ABE  is  a  right  angle;  therefore  EDA  is  also  a 
right  angle,  and  ED  perpendicular  to  DA:  but  it  is  also  perpen- 
dicular to  each  of  the  tvvo  BD,  DC:  wherefore  ED  is  at  right  an- 
gles to  each  of  the  three  straight  lines  BD,  DA,  DC  in  the  point 
in  which  they  meet:  therefore  these  three  straight  lines  are  all  in 
the  same  plane  (5.  1 1.):  but  AB  is  in  the  plane  in  which  are  BD, 
DA,  because  any  three  straight  lines  which  meet  one  another  are 
in  one  plane  (2.   11.):  therefore  AB,  BD,  DC  are  in  one  plane: 


BOOK  XI.  THE  ELEMENTS  OF  EUCLID.  163 

and  each  of  the  angles  ABD,  BDC  is  a  right  angle;  therefore 
AB  is  parallel  (28.  1.)  to  CD.  Wherefore,  if  two  straight  lines, 
&c.     Q.  E.  D. 

PROP.  VII.  THEOR. 

If  two  Straight  lines  be  parallel,  the  straight  line 
drawn  from  any  point  in  the  one  to  any  point  in  the 
other  is  in  the  same  plane  with  the  parallels.* 

Let  AB,  CD  be  parallel  straight  lines,  and  take  any  point  E  in 
the  one,  and  the  point  F  in  the  other:  the  straight  line  which 
joins  E  and  F  is  in  the  sanne  plane  with  the  parallels. 

If  not,  let  it  be,  if  possible,  above  the  plane,  as  EGF;  and  in  the 

plane  ABCD  in  which  the  parallels     A E B 

are,  draw  the  straight  line  EHF  from 

E  to  F;  and  since  EGF   also  is   a 

straight  line,  the  two  straight  lines 

EHF,  EGF  include  a  space  between 

them,  which  is  impossible  (10.  Ax. 

1.).      Therefore   the    straight    line     q  p  q 

joining  the  points  E,  F  is  not  above 

the  plane  in  which  the  parallels  AB,  CD  are,  and  is  therefore  in 

that  plane.     Wherefore,  if  two  straight  lines.  Sec.  Q.  E.  D. 

PROP.  VIII.  THEOR. 

If  two  straight  lines  be  parallel,  and  one  of  them  be 
at  right  angles  to  a  plane,  the  other  also  shall  be  at 
right  angles  to  the  same  plane.* 

Let  AB,  CD  be  two  parallel  straight  lines,  and  let  one  of  them 
AB  be  at  right  angles  to  a  plane:  the  other  CD  is  at  right  angles 
to  the  same  plane. 

Let  AB,  CD  meet  the  plane  in  the  points  B,  D,  and  join  BD: 
therefore  (7.  11.)  AB,  CD,  BD  are  in  one  plane.  In  the  plane  to 
which  AB  is  at  right  angles,  draw  DE  at  right  angles  to  I3D,  and 
make  DE  equal  to  AB,  and  join  BE,  AE,  AD.  And  because  AB 
is  perpendicular  to  the  plane,  it  is  perpendicular  to  every  straight 
line  which  meets  it,  and  is  in  that  plane  (3.  def.  11.);  therefore 
each  of  the  angles  ABD,  ABE  is  a  right  angle;  and  because  the 
straight  line  BD  meets  the  parallel  straight  lines  AB,  CD,  the 
angles  ABD,CDB  are  together  equal  (29.  1.)  to  two  right  angles: 
and  ABD  is  a  right  angle;  therefore  also  CDB  is  a  right  angle, 
and  CD  perpendicular  to  BD:  and  because  AB  is  equal  to  DE, 
and  BD  common,  the  two  AB,  BD  are  equal  to  the  two  ED,  DB, 
and  the  angle  ABD  is  equal  to  the  angle  EDB,  because  each  of 

*  See  Note, 


164 


THE  ELEMENTS  OF  EUCLID. 


BOOK   XI, 


D 


them  is  a  right  angle;  therefore  the  base  C 

AD  is  equal  (4.  1.)  to  the  base  BE:  again, 
because  AB  is  equal  to  DE,  and  EB  to 
AD;  the  two  AB,  BE  are  equal  to  the  two 
ED,  DA;  and  the  base  AE  is  common  to 
the  triangles  ABE,  EDA;  wherefore  the 
angle  ABE  is  equal  (8.  1.)  to  the  angle 
EDA,  and  ABE  is  a  right  angle;  and  there- 
fore EDA  is  a  right  angle,  and  ED  perpen- 
dicular to  DA:  but  it  is  also  perpendicular 
to  BD:  therefore  ED  is  perpendicular  (4. 
11.)  to  the  plane  which  passes  through 
BD,  DA,  and  shall  (3.  def.  1 1.)  make  right 
angles  with  every  straight  line  meeting  it 
in  that  plane:  but  DC  is  in  the  plane  passing  through  BD,  DA, 
because  all  three  are  in  the  plane  in  which  are  the  parallels  AB, 
CD;  wherefore  ED  is  at  right  angles  to  DC;  and  therefore  CD  is 
at  right  angles  to  DE:  but  CD  is  also  at  right  angles  to  DB; 
CD  then  is  at  right  angles  to  the  two  straight  lines  DE,  BD  in 
the  point  of  their  intersection  D;  and  therefore  is  at  right  angles 
(4.  11.)  to  the  plane  passing  through  DE,  DB,  which  is  the  same 
plane  to  which  AB  is  at  right  angles.  Therefore,  if  two  straight 
lines,  Sec.  Q.  E.  D. 

PROP.  IX.  THEOR. 

Two  Straight  lines  which  are  each  of  them  parallel 
to  the  same  straight  line,  and  not  in  the  same  plane 
with  it,  are  parallel  to  one  another. 

Let  AB,  CD  be  each  of  them  parallel  to  EF,  and  not  in  the 
same  plane  with  it;  AB  shall  be  parallel  to  CD. 

In  EF  take  any  point  G,  from  which  draw,  in  the  plane  passing 
through  EF,  AB,  the  straight  line  GH  at  right  angles  to  EF,  and 
in  the  plane  passing  through  EF,  CD,  draw  GK  at  right  angles  to 
the  same  EF.  And  because  EF  is  perpendicular  both  to  GH  and 
GK,  EF  is  perpendicular  (4.  11.)  AH  B 

to  the  plane  HGK  passing  through 
them:  and  EF  is  parallel  to  AB, 
therefore  AB  is  at  right  angles  (8. 
1 1.)  to  the  plane  HGK.  For  the 
same  reason,  CD  is  likewise  at 
right  angles  to  the  plane  HGK. 
Therefore,  AB,  CD  are  each  of 
them  at  right  angles  to  the  plane       C       K  D 

HGK.  But  if  two  straight  lines  be  at  right  angles  to  the  same 
plane,  they  shall  be  parallel  (6.  11.)  to  one  another.  Therefore 
AB  is  parallel  to  CD.    Wherefore,  two  straight  lines,  &:c.  Q.  E.  D. 


—  F 


PROP.  X.  THEOR. 


If  two  straight  lines  meeting  one  another  be  parallel 
to  two  other  that  meet  one  another,  and  are  not  in  the 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


165 


same  plane  with  the  first  two,  the  first  two  and  the 
other  two  shall  contain  equal  angles. 

Let  the  two  straight  lines  AB,  BC  which  meet  one  another  be 
parallel  to  the  two  straight  lines  DE,  EF  that  meet  one  another, 
and  are  not  in  the  same  plane  with  AB,  BC.  The  angle  ABC  is 
equal  to  the  angle  DEF. 

Take  BA,  BC,  ED,  EF  all  equal  to  one  another;  and  join  AD, 
CF,  BE,  AC,  DF;  because  BA  is  equal  and  parallel  to  ED,  there- 
fore AD  is  (33.  1.)  both  equal  and  parallel  B 
to  BE.  For  the  same  reason,  CF  is  equal 
and  parallel  to  BE.  Therefore  AD  and 
CF  are  each  of  them  equal  and  parallel  to 
BE.  But  straight  lines  that  are  parallel  to 
the  same  straight  line,  and  not  in  the  same 
plane  with  it,  are  parallel  (9.  11.)  to  one 
another.  Therefore  AD  is  parallel  to  CF; 
and  it  is  equal  (1.  Ax.  1.)  to  it,  and  AC, 
DF  join  them  towards  the  same  parts;  and 
therefore  (33.  1.)  AC  is  equal  and  parallel 
to  DF.  And  because  AB,  BC  are  equal  to 
DE,  EF,  and  the  base  AC  to  the  base  DF;  the  angle  ABC  is  equal 
(8.  1.)  to  the  angle  DEF.  Therefore,  if  two  straight  lines,  &c. 
Q.  E.  D. 


PROP.  XI.  PROB. 

To  draw  a  straight  line  perpendicular  to  a  plane^ 
from  a  given  point  above  it. 

Let  A  be  the  given  point  above  the  plane  BH;  it  is  required 
to  draw  from  the  point  A  a  straight  line  perpendicular  to  the 
plane  BH. 

In  the  plane  draw  any  straight  line  BC,  and  from  the  point  A 
draw  (12.  1.)  AD  perpendicular  to  BC.  If  then  AD  be  also  per- 
pendicular to  the  plane  BH,  the  thing  required  is  already  done; 
but  if  it  be  not,  from  the  point  D  draw  (11.  1.)  in  the  plane  BH, 
the  straight  lineDE  at  right  angles 
to  BC:  and  from  the  point  A  draw 
AF  perpendicular  to  DE;  and 
through  F  draw  (31.  1.)  GH  paral- 
lel to  BC:  and. because  BC  is  at 
right  angles  to  ED  and  DA:  BC  is 
at  right  angles  (4.  11.)  to  the  plane 
passing  through  ED,  DA.  And 
GH  is  parallel  to  BC;  but,  if  two 
straight  lines  be  parallel,  one  of 
which  is  at  right  angles  to  a  plane, 
the  other  shall  be  at  right  (8.  11.)  angles  to  the  same  plane; 
wherefore  GH  is  at  right  angles  to  the  plane  through  ED,  DA, 
and  is  perpendicular  (3.  def.  11.)  to  every  straight  line  meeting 
it   in  that  plane.     But  AF,  which  is  in  the  plane  through  ED, 


166  THE  ELEMENTS  OF  EUCLID.  BOOK  XI. 

DA,  meets  it:  therefore  GH  is  perpendicular  to  AF;  and  conse- 
quently AF  is  perpendicular  to  GH,  and  AF  is  perpendicular  to 
DE:  therefore  AF  is  perpendicular  to  each  of  the  straight  lines 
GH,  DE.  But  if  a  straight  line  stands  at  right  angles  to  each  of 
two  straight  lines  in  the  point  of  their  intersection,  it  shall  also 
be  at  right  angles  to  the  plane  passing  through  them.  But  the 
plane  passing  through  ED,  GH  is  the  plane  BH^  therefore  AF  is 
perpendicular  to  the  plane  BH^  therefore,  from  the  given  point 
A,  above  the  plane  BH,  the  straight  line  AF  is  drawn  perpendi- 
cular to  that  plane.     Which  was  to  be  done. 

PROP.  XH.  PROB. 

To  erect  a  straight  line  at  right  angles  to  a  given 
plane,  from  a  point  given  in  the  plane. 

Let  A  be  the  point  given  in  the  plane;  it  is  required  to  erect 
a  straight  line  from  the  point  A  at  right  an-  D      ~ 

gles  to  the  plane. 

From  any  point  B  above  the  plane  draw 
(11.  11.)  BC  perpendicular  to  it;  and  from 
A  draw  (31.  1.)  AD  parallel  to  BC.  Be- 
cause, therefore,  AD,  CB  are  two  parallel 
straight  lines,  and  one  of  them  BC  is  at 
right  angles  to  the  given  plane,  the  other 
AD  is  also  at  right  angles  to  it  (8.  11.). 
Therefore  a  straight  line  has  been  erected  at  right  angles  to  a 
given  plane  from  a  point  given  in  it.     Which  was  to  be  done. 

PROP.  XHI.  THEOR. 

From  the  same  point  in  a  given  plane,  there  cannot 
be  two  straight  lines  at  right  angles  to  the  plane,  upon 
the  same  side  of  it;  and  there  can  be  but  one  perpendi- 
cular to  a  plane  from  a  point  above  the  plane. 

For,  if  it  be  possible,  let  the  two  straight  lines  AC,  AB  be  at 
right  angles  to  a  given  plane  from  the  same  point  A  in  the  plane, 
and  upon  the  same  side  of  it:  and  let  a  plane  pass  through  BA, 
AC;  the  common  section  of  this  with  the  given  plane  is  a  straight 
(3.  11.)  line  passing  through  A:  let  DAE  be  their  common  sec- 
tion: therefore  the  straight  lines  AB,  AC,  DAE  are  in  one 
plane:  and  because  CA  is  at  right  angles  to  the  given  plane,  it 
shall  make  right  angles  with  every  straight  line  meeting  it  in  that 
plane.  But  DAE,  which  is  in  that  plane,  meets  CA;  therefore 
CAE  is  a  right  angle.     For  the  same  B  C 

reason  BAE  is  a  right  angle.  Where- 
fore the  angle  CAE  is  equal  to  the 
angle  BAE;  and  they  are  in  one 
plane,  which  is  impossible.  Also, 
from  a  point  above  a  plane,  there  can 
be  but  one  perpendicular  to  that  plane; 
for,  if  there  could  be  two,  they  would  ^ 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


\67 


K 


C 


/ 

/H\ 

\\ 

A 

IS 

E 
D 

\. 

be  parallel  (6.  11.)  to  one  another,  which  is  absurd.     Therefore, 
From  the  same  point,  8cc.     Q.  E.  D. 

PROP.  XIV.  THEOR. 

Planes  to  which  the  same  straight  hne  is  perpendi- 
cular, are  parallel  to  one  another. 

Let  the  straight  line  AB  be  perpendicular  to  each  of  the  planes 
CD,  EF;  these  planes  are  parallel  to  one  another. 

If  not,  they  shall  meet  one  another  when  produced;  let  them 
meet;  their  common  section  shall  be  a  G 

straight  line  GH,  in  which  take  any 
point  K,  and  join  AK,  BK:  then  be- 
cause AB  is  perpendicular  to  the  plane 
EF,  it  is  perpendicular  (3.  def.  1 1.)  to  the 
straight  line  BK  which  is  in  that  plane. 
Therefore  ABK  is  a  right  angle.  P'or 
the  same  reason,  BAK  is  a  right  angle; 
wherefore  the  two  angles  ABK,  BAK  of 
the  triangle  ABK  are  equal  to  two  right 
angles,  which  is  impossible  (17.  I.); 
therefore  the  planes  CD,  EF  though  pro- 
duced, do  not  meet  one  another;  that  is, 
they  are  parallel  (8.  def.  11.).     Therefore,  planes,^' Sec.     Q.  E.  D 

PROP.  XV.  THEOR. 

If  two  straight  lines  meeting  one  another,  be  parallel 
to  two  straight  lines  which  meet  one  another,  but  are 
not  in  the  same  plane  with  the  first  two,  the  plane  which 
passes  through  these  is  parallel  to  the  plane  passing  the 
others."^ 

Let  AB,  BC,  two  straight  lines  meeting  one  another,  be  paral- 
lel to  DE,  EF,  that  meet  one  another,  but  are  not  in  the  same 
plane  with  AB,  BC:  the  planes  through  AB,  BC,  and  DE,  EF 
shall  not  meet,  though  produced. 

From  the  point  B  draw  BG  perpendiclar  (I  I.  11.)  to  the  plane 
v/hich  passes  through  DE,  EF,  and  let  it  meet  that  plane  in  G; 
and  through  G  draw  GH  parallel  (31.  1.)  to  ED,  and  GK  paral- 
lel to  EF;  and  because  BG  is  perpendicular  to  the  plane  through 
DE,  EF,  it  shall  make  right  an- 
gles with  every  straight  line  meet- 
ing it  in  that  plane  (3.  def.  11.).  B 
But  the  straight  lines  GH,  GK 
in  that  plane  meet  it:  therefore 
each  of  the  angles  BGH,  BGK  is 
a  right  angle:  and  because  BA  is 
parallel  (9.  1 1.)  to  GH  (for  each  A 
of  them  is  parallel  to  DE,  and 
they  are  not  both    in  the  same 


See  Note. 


168 


THE    ELEMENTS    OF    EUCLID. 


BOOK  XI. 


plane  with  it)  the  angles  GBA,  BGH  are  together  equal  (29.  1.) 
to  two  right  angles:  and  BGH  is  a  right  angle,  therefore  als'o  GBA 
is  a  right  angle,  and  GB  perpendicular  to  BA^  for  the  same  rea- 
son, GB  is  perpendicular  to  BC:  since  therefore  the  straight  line 
GB  stands  at  right  angles  to  the  two  straight  lines  BA,  BC,  that 
cut  one  another  in  B^  GB  is  perpendicular  (4.  11.)  to  the  plane 
through  AB,  BC:  and  it  is  perpendicular  to  the  plane  through 
DE,  EF:  therefore  BG  is  perpendicular  to  each  of  the  planes 
through  AB,  BC,  and  DE,  EF:  but  planes  to  which  the  same 
straight  line  is  perpendicular,  are  parallel  (14.  11.)  to  one  another: 
therefore  the  plane  through  AB,  BC  is  parallel  to  the  plane 
through  DE,  EF.     Wherefore,  if  two  straight  lines  8cc.   Q.  E.  D. 

PROP.  XVI.  THEOR. 

If  two  parallel  planes  be  cut  by  another  plane,  their 
common  sections  with  it  are  parallels.* 

Let  the  parallel  planes  AB,  CD  be  cut  by  the  plane  EFHG, 
and  let  their  common  sections  with  it  be  EF,  GHj  EF  is  parallel 
to  GH. 

For,  if  it  be  not,  EF",  GFI  shall  meet,  if  produced,  either  on  the 
side  of  FH,or  EG;  first,  let  them  be  produced  on  the  side  of  FH, 
and  meet  in  the  point  K;  therefore,  since  EFK  is  in  the  plane  AB, 
every  point  in  EF'K  is  in  that  plane;  K 

and  K  is  a  point  in  EFK;  therefore 
K  is  in  the  plane  AB:  for  the  same 
reason  K  is  also  in  the  plane  CD: 
wherefore  the  planes  AB,  CD  pro- 
duced meet  one  another;  but  they 
do  not  meet,  since  they  are  paral- 
lel by  the  hypothesis:  therefore  the 
straight  lines  EF,  GH  do  not  meet 
when  produced  on  the  side  of  FH; 
in   the    same   manner   it    may   be 

proved,  that  EF,  GH  do  not  meet  ^   "^  G    "~' 

when  produced  on  the  side  of  EG:  but  straight  lines  which  are 
in  the  same  plane  and  do  not  meet,  though  produced  either  way, 
are  parallel:  therefore  EF  is  parallel  to  GH.  Wherefore,  if  two 
parallel  planes.  Sec.  Q.  E.  D. 

PROP.  XVH.  THEOR. 

If  two  straight  lines  be  cut  by  parallel  planes,  they 
shall  be  cut  in  the  same  ratio. 

Let  the  straight  lines  AB,  CD  be  cut  by  the  parallel  planes 
GH,  KL,  MN,  in  the  points  A,  E,  B;  C,  F,  D;  as  AE  is  to  EB, 
so  is  CF  to  FD. 

Join  AC,  BD,  AD,  and  let  AD  meet  the  plane  KL  in  the 
point  X:    and   join   EX,  XF:   because  the    two  parallel  planes 


"*  See  Note. 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


KL,  MN  are  cut  by  the  plane 
EX,  BD,  are  parallel  (16.  11.). 
the  two  parallel  planes  GH,  KL 
are  cut  by  the  plane  AXFC,  the 
common  sections  AC,  XF,  are 
parallel:  and  because  EX  is  pa- 
rallel to  BD,  a  side  of  the  triangle 
ABD,  as  AE  to  EB,  so  is  (2.  6.) 
AXtoXD.  Again,  because  XF 
is  parallel  to  AC,  a  side  of  the 
triangle  ABC,  as  AX  to  XD,  so 
is  CF  to  FD:  and  it  was  proved 
that  AX  is  to  XD,  as  AE  to 
EB;  therefore  (11.5.),  as  AE  to 
EB,  so  is  CF  to  FD.     Where- 


fore, if  two  straight  lines.  See.  Q.  E. 


EBDX,  the   common 
For  the  same  reason, 


169 

sections 
because 
H 


PROP.  XVIII.  THEOR. 

If  a  straight  line  be  at  right  angles  to  a  plane,  every 
plane  which  passes  through  it  shall  be  at  right  angles  to 
that  plane. 

Let  the  straight  line  AB  be  at  right  angles  to  a  plane  CK; 
every  plane  which  passes  through  AB  shall  be  at  right  angles  to 
the  plane  CK. 

Let  any  plane  DE  pass  through  AB,  and  let  CE  be  the  com- 
mon section  of  the  planes  DE,  CK;  take  any  point  F  in  CE, 


D 


G 


H 


K 

\ 

N 

from  which  draw  FG  in  the  plane 
DE  at  right  angles  to  CE;  and 
because  AB  is  perpendicular  to 
the  plane  CK,  therefore  it  is  also 
perpendicular  to  every  straight 
line  in  that  plane  meeting  it  (3. 
def.  11.);  and  consequently  it  is 
perpendicular  to  CE:  wherefore 
ABF  is  a  right  angle;  but  GFB 
is  likewise  a  right  angle:  there-  C  F  B         E 

fore  AB  is  parallel  (28.  1.)  to  FG.  And  AB  is  at  right  angles  to 
the  plane  CK:  therefore  FG  is  also  at  right  angles  to  the  same 
plane  (8.  11.).  But  one  plane  is  at  right  angles  to  another  plane 
when  the  straight  lines  drawn  in  one  of  the  planes  at  right  angles 
to  their  common  section,  are  also  at  right  angles  to  the  other 
plane  (4.  def.  11.):  and  any  straight  line  FG  in  the  plane  DE, 
which  is  at  right  angles  to  CE  the  common  section  of  the  planes, 
has  been  proved  to  be  perpendicular  to  the  other  plane  CK;  there- 
fore the  plane  DE  is  at  right  angles  to  the  plane  CK.  In  like 
manner,  it  may  be  proved  that  all  the  planes  which  pass  through 
AB  are  at  right  angles  to  the  plane  CK.  Therefore,  if  a  straight 
line.  See.  Q.  E.  D. 
Y 


170 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI. 


PROP.  XIX.  TPIEOR. 

If  two  planes  cutting  one  another  be  each  of  them 
perpendicular  to  a  third  plane:  their  common  section 
shall  be  perpendicular  to  the  same  plane. 

Let  the  two  planes  AB,  BC  be  each  of  them  perpendicular  to 
a  third  plane,  and  let  BD  be  the  common  section  of  the  first  twoj 
BD  is  perpendicular  to  the  third  plane. 

If  it  be  not,  from  the  point  D  draw  in  the  plane  AB,  the 
straight  line  DE  at  right  angles  to  AD,  the  common  section  of 
the  plane  AB  with  the  third  plane;  and  in  the  plane  BC  draw 
DF  at  right  angles  to  CD  the  common  section  of  the  plane  BC 
with  the  third  plane.  And  because  the  plane 
AB  is  perpendicular  to  the  third  plane, 
and  DE  is  drawn  in  the  plane  AB  at  right 
angles  to  AD  their  common  section,  DE  is  ^ 
perpendicular  to  the  third  plane  (4.  def.  1 1.). 
In  the  same  manner,  it  may  be  proved  that 
DF  is  perpendicular  to  the  third  plane. 
Wherefore,  from  the  point  D  two  straight 
lines  stand  at  right  angles  to  the  third  plane, 
upon  the  same  side  of  it,  which  is  impossi- 
ble (13.  11.):  therefore,  from  the  point  D 
there  cannot  be  any  straight  line  at  right 
angles  to  the  third  plane,  except  BD  the  com-     A  C 

mon  section  of  the  planes  AB,  BC.     BD  therefore  is  perpendicu- 
lar to  the  third  plane.     Wherefore,  if  two  planes.  Sec.    Q.  E.  D. 


PROP.  XX.  THEOR. 

If  a  solid  angle  be  contained  by  three  plane  angles, 
any  two  of  them  are  greater  than  the  third.^ 

Let  the  solid  angle  at  A  be  contained  by  the  three  plane  an- 
gles, BAC,  CAD,  DAB.  Any  two  of  them  are  greater  than  the 
third. 

If  the  angles  BAC,  CAD,  DAB  be  all  equal,  it  is  evident  that 
any  two  of  them  are  greater  than  the  third.  But  if  they  be  not, 
let  BAC  be  that  ano-le  which  is  not  less  than  either  of  the  other 
two,  and  is  greater  than  one  of  them  DAB;  and  at  the  point  A  in 
the  straight  line  AB,  make,  in  the  plane  which  passes  through 
BA,  AC,  the  angle  BAE  equal  (23.  L)  to  the  angle  DAB;  and 
make  AE  equal  to  AD,  and  through  E  draw  BEC  cutting  AB, 
AC  in  the  points  B,  C,  and  join  DB,  DC.     And  because  DA  is 


See  Note 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


A 


equal  to  AE,  and  AB  is  common,  the  d 

two  DA,  AB  are  equal  to  the  two  EA, 
AB,  and  the  angle  DAB  is  equal  to  the 
angle  EAB:  therefore  the  base  DB  is 
equal  (4.  1.)  to  the  base  BE.  And  be- 
cause BD,  DC  are  greater  (20.  1.)  than 
CB,  and  one  of  them,  BD,  has  been 
proved  equal  to  BE  a  part  of  CB,  there- 
fore the  other  DC  is  greater  than  the 
remaining  part  EC.  And  because  DA  is  equal  to  EA''^  and  AC 
common,  but  the  base  DC  greater  than  the  base  EC:  therefore 
the  angle  DAC  is  greater  (25.  1.)  than  the  angle  EAC:  and,  by 
the  construction,  the  angle  DAB  is  equal  to  the  angle  BAE; 
wherefore  the  angles  DAB,  DAC  are  together  greater  than  BAE, 
EAC,  that  is,  than  the  angle  BAC.  But  BAC  is  not  less  than 
either  of  the  angles  DAB,  BAC:  therefore  BAC,  with  either  of 
them,  is  greater  than  the  other.  Wherefore,  if  a  solid  angle,  &c. 
Q.  E.  D. 

PROP.  XXI.  THEOR. 

Every  solid  angle  is  contained  by  plane  angles  which 
together  are  less  than  four  right  angles. 

First,  let  the  solid  angle  at  A  be  contained  by  three  plane  angles 
BAC,  CAD,  DAB.  These  three  together  are  less  than  four  right 
angles. 

Take  in  each  of  the  straight  lines  AB,  AC,  AD  any  points  B, 
C,  D,  and  join  BC,  CD,  DB:  then,  because  the  solid  angle  at  B, 
is  contained  by  the  three  plane  angles  CBA,  ABD,  DBC,  any  two 
of  them  are  greater  (20.  11.)  than  the  third;  therefore  the  angles 
CBA,  ABD,  are  greater  than  the  angle  DBC;  for  the  same  rea- 
son, the  angles  BCA,  ACD  are  greater  than  the  angle  DCB;  and 
the  angles  CDA,  ADB,  greater  than  BDC:  wherefoie  the  six 
angles  CBA,  ABD,  BCA,  ACD,  CDA,  ADB,  are  greater  than 
the  three  angles  DBC,  BCD,  CDB;  but  D 

the  three  angles  DBC,  BCD,  CDB,  are 
equal  to  two  right  angles  (32.  1.):  there- 
fore the  six  angles  CBA,  ABD,  BCA, 
ACD,  CDA,  ADB  are  greater  than  two 
right  angles:  and  because  the  three  an- 
gles of  each  of  the  triangles  ABC,  ACD, 
ADB  are  equal  to  two  right  angles, 
therefore  the  nine  angles  of  these  three  triangles,  viz.  the  angles 
CBA,  BAC,  ACB,  ACD,  CDA,  DAC,  ADB,  DBA,  BAD  are 
equal  to  six  right  angles;  of  these  the  six  angles  CBA,  ACB, 
ACD,  CDA,  ADB,  DBA  are  greater  than  two  light  angles: 
therefore  the  remaining  three  angles  BAC,  DAC,  BAD,  which 
contain  the  solid  angle  at  A,  are  less  than  four  right  angles. 

Next,  let  the  solid  angle  at  A  be  contained  by  any  number  of 
plane  angles  BAC,  CAD,  DAE,  EAF,  FAB;  these  together  are 
less  than  four  right  angles. 

Let  the  planes  in  which  the  angles  are,  be  cut  by  a  plane,  and 


172  THE  ELEMENTS  OF  EUCLID.  BOOK  XI. 

let  the  common  section  of  it  with  those  planes  be  BC,  CD,  DE, 
EF,  FB;  and  because  the  solid  angle  at  B  is  contained  by  three 
plane  angles  CBA,  ABF,  FBC,  of  which   any  two  are  greater 
(20.  11.)  than  the  third,  the  angles  CBA,  ABF,  are  greater  than 
the  angle  FBC:  for  the  same  reason,  the 
two  plane  angles  at  each  of  the  points  C, 
D,  E,  F,  viz.  the  angles  which  are  at  the 
bases  of  the  triangles,  having  the  com- 
mon  vertex  A,   are    greater   than   the 
third  angle  at  the  same  point,  which  is 
one  of  the  angles  of  the  polygon  BCDEF: 
therefore  all  the  angles  at  the  bases  of 
the  triangles  are  together  greater  than 
all  the  angles  of  the  polygon:  and  be- 
cause all  the  angles  of  the  triangles  are 
together  equal  to  twice  as  many  right 
angles  as  there   are  triangles  (32.  I.); 
that  is,  as  there  are  sides  in  the  polygon  BCDEF:  and  that  all 
the  angles  of  the  polygon,  together  with  four  right  angles,  are 
likewise  equal  to  twice  as  many  right  angles  as  there  are  sides  in 
the  polygon  (l.  Cor.  32.  1.):  therefore  all  the  angles  of  the  trian- 
gles are  equal  to  all  the  angles  of  the  polygon  together  with  four 
right  angles.     But  all  the  angles  at  the  bases  of  the  triangles  are 
greater  than  all  the  angles  of  the  polygon,  as  has  been  proved. 
Wherefore,  the  remaining  angles  of  the  triangles,  viz.  those  at  the 
vertex,  which  contain  the  solid  angle  at  A,  are  less  than  four 
right  angles.     Therefore  every  solid  angle,  Sec.  Q.  E.  D. 

PROP.  XXII.  THEOR. 

If  every  two  or  three  plane  angles  be  greater  than 
the  third,  and  if  the  straight  lines  which  contain  them 
be  all  equal;  a  triangle  may  be  made  of  the  straight 
lines,  that  join  the  extremities  of  those  equal  straight 
lines."^ 

Let  ABC,  DEF,  GHK  be  three  plane  angles,  whereof  every 
two  are  greater  than  the  third,  and  are  contained  by  the  equal 
straight  lines  AB,  BC,  DE,  EF,  GH,  HK;  if  their  extremities  be 
joined  by  the  straight  lines  AC,  DF,  GK,  a  triangle  may  be  made 
of  three  straight  lines  equal  to  AC,  DF,  GK;  that  is,  every  two 
of  them  together  greater  than  the  third. 

If  the  angles  at  B,  E,  H  are  equal;  AC,  DF,  GK  are  also  equal 
(4.  1.),  and  any  two  of  them  greater  than  the  third:  but  if  the 
angles  be  not  all  equal,  let  the  angle  ABC  be  not  less  than  either 
of  the  two  at  E,  H;  therefore  the  straight  line  AC  is  not  less  than 
either  of  the  other  two  DF,  GK  (4.  Cor.  24.  1.);  and  it  is  plain 
that  AC,  together  with  either  of  the  other  two,  must  be  greater 
than  the  third:  also,  DF,  with  GK,  are  greater  than  AC:  for  at 
the  point  B  in  the  straight  line  AB  make  (23.  1.)  the  angle  ABL 

*  See  Note. 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


173 


equal  to  the  angle  GHK,  and  tnake  BL  equal  to  one  of  the 
straight  lines  AB,  BC,  DE,  EF,  GH,  HK,  and  join  AL,  LC,-  then 
because  AB,  BL  are  equal  to  GH,  HK,  and  the  angle  ABL  to  the 
angle  GHK,  the  base  AL  is  equal  to  the  base  GK;  and  because 
the  angles  at  E,  H  are  greater  than  the  angle  ABC,  of  which  the 
angle  at  H  is  equal  to  ABL;  therefore  the  remaining  angle  at  E 
is  greater  than  the  angle  LBC^  and  because  the  two  sides  LB, 

B 


D 


F     c; 


Be  are  equal  to  the  two  DE,  EF,  and  that  the  angle  DEF  is 
is  greater  than  the  angle  LBC,  the  base  DF  is  greater  (24.  1.)  than 
the  base  LC:  and  it  has  been  proved  that  GK  is  equal  to  AL; 
therefore  DF  and  GK  are  greater  than  AL  and  LC;  but  AL  and 
LC  are  greater  (20.  1.)  than  AC:  much  more  then  are  DF  and 
GK  greater  than  AC.  Wherefore  every  two  of  these  straight 
lines  AC,  DF,  GK  are  greater  than  the  third;  and,  therefore,  a  tri- 
angle may  be  made  (22.  1.)  the  sides  of  which  shall  be  equal  to 
AC,DF,  GK.     Q.  E.  D. 

PROP.  XXm.  PROB. 

To  make  a  solid  angle  which  shall  be  contained  by 
three  given  plane  angles,  any  two  of  them  being 
greater  than  the  third,  and  all  three  together  less  than 
four  right  angles.* 

Let  the  three  given  plane  angles  be  ABC,  DEF,  GHK,  any  two 
of  which  are  greater  than  the  third,  and  all  of  them  together  less 
than  four  right  angles.  It  is  required  to  make  a  solid  angle  con- 
tained by  three  plane  angles  equal  to  ABC,  DEF,  GHK,  each  to 
each. 

B  H 


D 


*  See  Note. 


174 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI, 


From  the  straight  lines  containing  the  angles,  cut  off  AB,  BC, 
DE,  EF,  GH,  HK,  all  equal  to  one  another;  and  join  AC,  DF, 
GK;  then  a  triangle  may  be  made  (22.  11.)  of  three  straight  lines 
equal  to  AC,  DF,  GK.  Let  this  be  the  triangle  LMN  (22.  1.)  so 
that  AC  be  equal  to  LM,  DF  to  MN,  and  GK  to  LN;  and  about 
the  triangle  LMN  describe  (5.  4.)  a  circle,  and  find  its  centre  X, 
which  will  either  be  within  the  triangle,  or  in  one  of  its  sides,  or 
without  it. 

First;  let  the  centre  X  be  within  the  triangle,  and  join  LX, 
MX,  NX:  AB  is  greater  than  LX;  if  not,  AB  must  either  be  equal 
to,  or  less,  than  LX;  first,  let  it  be  equal:  then  because  AB  is  equal 
to  LX,  and  that  AB  is  also  equal  to  BC,  and  LX  to  XM,  AB  and 
BC  are  equaj  to  LX  and  XM,  each  to  each;  and  the  base  AC 
is,  by  construction,  equal  to  the  base  LM:  wherefore  the  angle 
ABC-is  equal  to  the  angle  LXM  (8.  1.).  For  the  same  reason, 
jLhfe'  angle  DEF  is  equal  to  the  angle  R 

MXN,  and  the  angle  GHK  to  the 
angle  NXL;  therefore  the  three  an- 
gles ABC,  DEF,  GHK  are  equal  to 
the  three  angles  LXM,  MXN, 
NXL:  but  the  three  angles  LXM, 
MXN,  NXL  are  equal  to  four  right 
angles  (2.  Cor.  15.  1.):  therefore  also 
the  three  angles  ABC,  DEF,  GHK, 
are  equal  to'four  right  angles;  but 
by  the  hypothesis,  they  are  less  than 
four  right  angles,  which  is  absurd; 
therefore  AB  is  not  equal  to  LX: 
but  neither  can  AB  be  less  than 
LX:  for,  if  possible,  let  it  be  less,  and  upon  the  straight  line  LM, 
the  side  of  it  on  which  is  the  centre  X,  describe  the  triangle  LOM, 
the  sides  LO,  OM  of  which  are  equal  to  AB,  BC;  and  because 
the  base  LM  is  equal  to  the  base  AC,  the  angle  LOM  is  equal  to 
the  angle  ABC  (8.  1.):  and  AB,  that  is  LO,  by  the  hypothesis, 
is  less  than  LX;  wherefore  LO,  OM  fall  within  the  triangle  LXM; 
for  if  they  fell  upon  its  sides,  or  without  it,  they  would  be  equal  to, 
or  greater  than  LX,  XM  (21.  1.):  therefore  the  angle  LOMy  that 
is  the  angle  ABC,  is  greater  than  the  angle  LXM  (21.  1.):  in  the 
same  manner  it  may  be  proved  that  the 
angle  DEF  is  greater  than  the  angle 
MXN,  and  the  angle  GHK  greater  than 
the  angle  NXL.  Therefore  the  three 
angles  ABC,  DEF,  GHK  are  greater 


R 


than  the  three 


angles 


LXM,  MXN, 


1  • 
•3? 


NXL;  that  is,  than  four  right  angle 
but  the  same  angles  ABC,  DEF,  GHK 
are  less  than  four  right  angles;  which 
is  absurd:  therefore  AB  is  not  less  than 
LX,  and  it  has  been  proved  that  it  is 
not  equal  to  LX;  wherefore  AB  is 
greater  than  LX. 

Next,  let  the  centre  X  of  the  circle 


M\ 


BOOK  XI, 


THE  ELEMENTS  OF  EUCLID. 


175 


/'  '\ 


fall  in  one  of  the  sides  of  the  triangle,  viz.  in  MN,  and  join  XL:  in 
this  case  also  AB  is  greater  than  LX.  If  not,  AB  is  either  equal 
to  LX,  or  less  than  it;  first,  let  it  be  equal  to  XL:  therefore  AB 
and   BC,  that  is,  DE  and  EF,  are  R 

equal  to  MX  and  XL,  that  is,  to  MN: 
but  by  the  construction,  MN  is  equal 
to  DF;  therefore  DE,  EF  are  equal 
to  DF,  which  is  impossible  (20.  1.): 
wherefore  AB  is  not  equal  to  LX; 
nor  is  it  less;  for  then,  much  more, 
an  absurdity  would  follow:  therefore 
AB  is  greater  than  LX. 

But  let  the  centre  X  of  the  circle 
fall  without  the  triangle  LMN,  and 
join  LX,  MX,  NX.  In  this  case 
likewise  AB  is  greater  than  LX:  if 
not,  it  is  either  equal  to  or  less  than  LX:  first,  let  it  be  equal;  it 
may  be  proved  in  the  same  manner,  as  in  the  first  case,  that  the 
angle  ABC  is  equal  to  the  angle  MXL,  and  GHK  to  LXN;  there- 
fore the  whole  angle  MXN  is  equal  to  the  two  angles  ABC,  GHK; 
but  ABC  and  GHK  are  together  greater  than  the  angle  DEF; 
therefore  also  the  angle  MXN  is  greater  than  DEF.  And  because 
DE,  EF  are  equal  to  MX,  XN,  and  the  base  DF  to  the  base  MN, 
the  angle  MXN  is  equal  (8.  1.)  to  the  angle  DEF:  and  it  has  been 
proved  that  it  is  greater  than  DEF,  which  is  absurd.  Therefore 
AB  is  not  equal  to  LX.  Nor  yet  is  it  less;  for  then,  as  has  been 
proved  in  the  first  case,  the  angle  ABC  is  greater  than  the  angle 
MXL,  and  the  angle  GHK  greater  than  the  angle  LXN.  At  the 
B  H 


point  B  in  the  straight  line  CB  make  the  angle  CBP  equal  to  the 
angle  GHK,  and  make  BP  equal  to  HK,  and  join  CP,  AP.  And 
because  CB  is  equal  to  GH;  CB,  BP  are  equal  to  GH,  HK,  each 
to  each,  and  they  contain  equal  angles;  wherefore  the  base  CP  is 
equal  to  the  base  GK,  that  is,  to  LN.  And  in  the  isosceles  tri- 
angles ABC,  MXL,  because  the  angle  ABC  is  greater  than  the 
angle  MXL,  therefore  the  angle  MLXat  the  base  is  greater  (32.  1.) 
than  the  angle  ACB  at  the  base.  For  the  same  reason,  because 
the  angle  GHK,  or  CBP,  is  greater  than  the  angle  LXN,  the 
angle  XLN  is  greater  than  the  angle  CBP.  Therefore  the  whole 
angle  MEN  is  greater  than  the  whole  angle  ACP.  And  because 
ML,  LN  are  equal  to  AC,  CP,  each  to  each,  but  the  angle  MLN  is 


176 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI. 


greater  than  the  angle  ACP,  the  base  R 

MN  is  greater  (24.  1.)  than  the  base 
AP.  And  MN  is  equal  to  DF,- 
therefore  also  DF  is  greater  than 
AP.  Again,  because  DE,  EF  are 
equal  to  AB,  BP,  but  the  base  DF 
greater  than  the  base  AP,  the  angle 
DEF  is  greater  (25.  1.)  than  the  an- 
gle ABP.  And  ABP  is  equal  to  the 
two  angles  ABC,  CBP,  that  is,  to  the  M 
two  angles  ABC,  GHK;  therefore 
the  angle  DEF  is  greater  than  the 
two  angles  ABC,  GHK;  but  it  is  also 
less  than  these,  which  is  impossible. 
Therefore  AB  is  not  less  than  LX, 
and  it  has  been  proved  that  it  is  not  equal  to  it^  therefore  AB  is 
greater  than  LX. 

From  the  point  X  erect  (12.  11.)  XR  at  right  angles  to  the 
plane  of  the  circle  LMN.  And  because  it  has  been  proved  in  all 
the  cases,  that  AB  is  greater  than  LX,  find  a  square  equal  to  the 
excess  of  the  square  of  AB  above  the 
square  of  LX,  and  make  RX  equal  to 
its  side,*  and  join  RL,  RM,  RN.  Be- 
cause RX  is  perpendicular  to  the  plane 
of  the  circle  LMN,  it  is  (3.  def.  11.) 
perpendicular  to  each  of  the  straight 
lines  LX,  MX,  NX.  And  because 
LX  is  equal  to  MX,  and  XR  common, 
and  at  right  angles  to  each  of  them, 
the  base  RL  is  equal  to  the  base  RM. 
For  the  same  reason,  RN  is  equal 
to  each  of  the  two  RL,  RM.  There- 
fore the  three  straight  lines  RL,  RM, 
RN  are  all  equal.  And  because  the 
square  of  XR  is  equal  to  the  excess 
of  the  square  of  AB  above  the  square**  of  LXj  therefore  the  square 
of  AB  is  equal  to  the  squares  of  LX,  XR.  But  the  square  of 
RL  is  equal  (47.  1.)  to  the  same  squares,  because  LXR  is  a  right 
angle.  Therefore  the  square  of  AB  is  equal  to  the  square  of  RL, 
and  the  straight  line  AB  to  RL.  But  each  of  the  straight  lines 
BC,  DE,  EF,  GH,  HK  is  equal  to  AB,  and  each  of  the  two  RM, 
RN  is  equal  to  RL.  Wherefore  AB,  BC,  DE,  EF,  GH,  HK  are 
each  of  them  equal  to  each  of  the  straight  lines  RL,  RM,  RN. 
A^i^  because  RL,  RM  are  equal  to  AB,  BC,  and  the  base  LM  to 
the  base  AC;  the  angle  LRM  is  equal  (8.  I.)  to  the  angle  ABC. 
For  the  same  reason,  the  angle  MRN  is  equal  to  the  angle  DEF, 
and  NRL  to  GHK.  Therefore  there  is  made  a  solid  angle  at  R, 
which  is  contained  by  three  plane  angles  LRM,  MRN,  NRL, 
which  are  equal  to  the  three  given  plane  angles  ABC,  DEF,  GHK, 
each  to  each.     Which  was  to  be  done. 


N 


BOOK  XI.  THE  ELEMENTS  OF  EUCLID.  177 

PROP.  A.    THEOR. 

If  each  of  two  solid  angles  be  contained  by  three  plane 
angles  equal  to  one  another,  each  to  each;  the  planes  in 
which  the  equal  angles  are,  have  the  same  inclination  to 
one  another.^ 

Let  there  be  two  solid  angles  at  the  points  A,  B;  and  let  the  angle 
at  A  be  contained  by  the  three  plane  angles  CAD,  CAE,  EAD; 
and  the  angle  at  B  by  the  three  plane  angles  FBG,  FBH,  HBG, 
of  which  the  angle  CAD  is  equal  to  the  angle  FBG;  and  CAE 
to  FBH;  and  EAD  to  HBG:  the  planes  in  which  the  equal  angles 
are,  have  the  same  inclination  to  one  another. 

In  the  straight  line  AC  take  any  point  K,  and  in  the  plane 
CAD  from  K  draw  the  straight  line  KD  at  right  angles  to  AC, 
and  in  the  plane  CAE  A  B 

the  straight  line  KL  at 
right  angles  to  the  same 
AC:  therefore  the  angle 
DKL  is  the  inclination 
(9.  def.  1 1 .)  of  the  plane 
CAD  to  the  plane  CAE. 
In  BF  take  BM  equal  to 
AK,  and  from  the  point 
M  draw,  in  the  planes 
FBG,  FBH,  the  straight  lines  MG,  MN  at  right  angles  to  BF; 
therefore  the  angle  GMN  is  the  inclination  (6.  def.  11.)  of  the 
plane  FBG  to  the  plane  FBH;  join  LD,  NG;  and  because  in  the 
triangles  KAD,  MBG,  the  angles  KAD,  MBG  are  equal,  as  also 
the  right  angles  AKD,  BMG,  and  that  the  sides  AK,  BM,  adja- 
cent to  the  equal  angles,  are  equal  to  one  another;  therefore  KD 
is  equal  (26.  1.)  to  MG,  and  AD  to  BG:  for  the  same  reason,  in 
the  triangles  KAL,  MBN,  KL  is  equal  to  MN,  and  AL  to  BN: 
and  in  the  triangles  LAD,  NBG,  LA,  AD  are  equal  to  NB,  BG, 
arid  they  contain  equal  angles;  therefore  the  base  LD  is  equal  (4. 
1.)  to  the  base  NG.  Lastly,  in  the  triangles  KLD,  MNG,  the 
sides  DK,  KL  are  equal  to  GM,  MN,  and  the  base  LD  to  the  base 
NG:  therefore  the  angle  DKL  is  equal  (8.  1.)  to  the  angle  GMN: 
but  the  angle  DKL  is  the  inclination  of  the  plane  CAD  to  the 
plane  CAE,  and  the  angle  GMN  is  the  inclination  of  the  plane 
FBG  to  the  plane  FBH,  which  planes  have  therefore  the  same 
inclination  (7.  def.  11.)  to  one  another:  and  in  the  same  manner 
it  may  be  demonstrated,  that  the  other  planes  in  which  the  equal 
angles  are,  have  the  same  inclination  to  one  another.  Therefore, 
if  two  solid  angles,  8cc.     Q.  E.  D. 

PROP.  B.  THEOR. 

If  two  solid  angles  be  contained,  each  by  three  plane 
angles  which  are  equal  to  one  another,  each  to  each,  and 

**  See  Note, 
Z 


1/8  THE  ELEMENTS  OF  EUCLID.  BOOK  XI,* 

alike  situated ;  these  solid  angles  are  equal  to  one  an- 
other."^ 

Let  there  be  two  solid  angles  at  A  and  B,  of  which  the  solid 
angle  at  A  is  contained  by  the  three  plane  angles  CAD;  CAE, 
EAD;  and  that  at  B,  by  the  three  plane  angles  FBG,  FBH,  HBG5 
of  which  CAD  is  equal  to  FBG;  CAE  to  FBH;  and  EAD  to 
HBG:  the  solid  angle  at  A  is  equal  to  the  solid  angle  at  B. 

Let  the  solid  angle  at  A  be  applied  to  the  solid  angle  at  B; 
and,  first,  the  plane  angle  CAD  being  applied  to  the  plane  angle 
FBG,  so  as  the  point  A  may  coincide  with  the  point  B,  and  the 
straight  line  AC  with  BF;  then  AD  coincides  with  BG,  be- 
cause the  angle  CAD  is  equal  to  the  *  ^ 
angle  FBG;  and  because  the  inclina- 
tion of  the  plane  CAE  to  the  plane 
CAD  is  equal  (A.  ll.)  to  the  inclina- 
tion of  the  plane  FBH  to  the  plane 
FBG,  the  plane  CAE  coincides  with 
the  plane  FBH,  because  the  planes 
CAD,  FBG  coincide  with  one  another:  and  because  the  straight 
lines  AC,  BF  coincide,  and  that  the  angle  CAE  is  equal  to  the 
angle  FBH;  therefore  AE  coincides  with  BH,  and  AD  coincides 
with  BG;  wherefore  the  plane  EAD  coincides  with  the  plane 
HBG:  therefore  the  solid  angle  A  coincides  with  the  solid  angle 
B,  and  consequently  they  are  equal  (A.  8.  1.)  to  one  another. 
Q.  E.  D. 

PROP.  C.  THEOR. 

Solid  figures  contained  by  the  same  number  of  equal 
and  similar  planes  alike  situated,  and  having  none  of 
their  solid  angles  contained  by  more  than  three  plane 
angles;  are  equal  and  similar  to  one  another.^ 

Let  AG,  KQ  be  two  solid  figures  contained  by  the  same  num- 
ber of  similar  and  equal  planes,  alike  situated,  viz.  let  the  plane 
AC  be  similar  and  equal  to  the  plane  KM;  the  plane  AF  to  KP; 
BG  to  LQ;  GD  to  QN;  DE  to  NO;  and  lastly,  FH  similar  and 
equal  to  PR:  the  solid  figure  AG  is  equal  and  similar  to  the  solid 
figure  KQ. 

Because  the  solid  angle  at  A  is  contained  by  the  three  plane  an- 
gles BAD,  BAE,  EAD,  which,  by  the  hypothesis,  are  equal  to 
the  plane  angles  LKN,  LKO,  OKN,  which  contain  the  solid  angle 
at  K,  each  to  each;  therefore  the  solid  angle  at  A  is  equal  (B.  11.) 
to  the  solid  angle  at  K:  in  the  same  manner,  the  otlier  solid  angles 
of  the  figures  are  equal  to  one  another.  If,  then,  the  solid  figure 
AG  be  applied  to  the  solid  figure  KQ,  first,  the  plane  figure  AC 

*  See  Note. 


BOOK   XI. 

being  applied  to  the 
plane  figure  KM: 
the  straight  line  AB 
coinciding  with  KL 
the  figure  AC  must 
coincide  with  the 
figure  KM,  because 
they  are  equal  and 


THE   ELEMENTS    OF    EUCLID. 


179 


H 


G 


R 


D 


K 


o 


Q 


> 


Jm! 


^^i 


K 


similar:  therefore  the  straight  lines  AD,  DC,  CB,  coincide  with 
KN,  NM,  ML,  each  with  eachj  and  the  points  A,  D,  C,  B,  with 
the  points  K,  N,  M,  L:  and  the  solid  angle  at  A  coincides  with 
(B.  11.)  the  solid  angle  at  K;  wherefore  the  plane  AF  coincides 
with  the  plane  KP,  and  the  figure  AF  with  the  figure  KP,  because 
they  are  equal  and  similar  to  one  another:  therefore  the  straight 
lines  AE,  EF,  FB,  coincide  with  KO,  OP,  PL;  and  the  points  E, 
F,  with  the  points  O,  P.     In  the  same  manner,  the  figure  AH  co- 
incides with  the  figure  KR,  and  the  straight  line  DH  with  NR, 
and  the  point  H  with  the  point  R:  and  because  the  solid  angle  at 
B  is  equal  to  the  solid  angle  at  L,  it  may  be  proved,  in  the  same 
manner,  that  the  figure  BG  coincides  with  the  figure  LQ,  and  the 
straight  line  CG  with  MQ,  and  the  point  G  with  the  point  Q: 
since,  therefore,  all  the  planes  and  sides  of  the  solid  figure  AG 
coincide  with  the  planes  and  sides  of  the  solid  figure  KQ,  AG  is 
equal  and  similar  to  KQ:  and,  in  the  same  manner,  any  other  solid 
figures  whatever  contained  by  the  same  number  of  equal  and  si- 
milar planes,  alike  situated,  and  having  none  of  their  solid  angles 
contained  by  more  than  three  plane  angles,  may  be  proved  to  be 
equal  and  similar  to  one  another.     Q.  E.  D. 

PROP.  XXIV.  THEOR. 

If  a  solid  be  contained  by  six  planes,  two  and  two  of 
which  are  parallel;  the  opposite  planes  are  similar  and 
equal  parallelograms."^ 

Let  the  solid  CDGH  be  contained  by  the  parallel  planes  AC, 
GF;  BG,  CE;  FB,  AE:  its  opposite  planes  are  similar  and  equal 
parallelograms. 

Because  the  parallel  planes  BG,  CE  are  cut  by  the  plane  AC, 
their  common  sections  AB,  CD  are  parallel  (16.  11.).  Again, 
because  the  two  parallel  planes  BF,  AE  are  cut  by  the  plane  AC, 
their  common  section  AD,  BC  are  parallel  (16.  11.);  and  AB  is 
parallel  to  CD;  therefore  AC  is  a  parallelogram.  In  like  man- 
ner, it  may  be  proved  that  each  of  the 
figures  CE,  FG,  GB,  BF,  AE  is  a  pa- 
rallelogram: join  AH,  DF;  and  be- 
cause AB  is  parallel  to  DC,  and  BH 
to  CF;  the  two  straight  lines  AB,  BH, 
which  meet  one  another,  are  parallel 
to  DC  and  CF  which  meet  one  ano- 
ther, and  are  not  in  the  same  plane 
with  the  other  two:  wherefore  they 
contain  equal  angles  (10.  II.);  the  an- 

*  See  Note. 


B 


H 


180 


rHE  ELEMExNTS  OF  EUCLID. 


BOOK  XI. 


gle  ABH  is  therefore  equal  to  the  angle  DCF;  and  because  AB, 
BH  are  equal  to  DC,  CF,  and  the  angle  ABH  equal  to  the  angle 
DCF;  therefore  the  base  AH  is  equal  (4.  1.)  to  the  base  DF,  and 
the  triangle  ABH  to  the  triangle  DCF:  and  the  parallelogram  BG 
is  double  (34.  1.)  of  the  triangle  ABH,  and  the  parallelogram  CE 
double  of  the  triangle  DCF;  therefore  the  parallelogram  BG  is 
equal  and  similar  to  the  parallelogram  CE.  In  the  same  manner 
it  may  be  proved,  that  the  parallelogram  AC  is  equal  and  similar 
to  the  parallelogram  GF,  and  the  parallelogram  AE  to  BF.  There- 
fore, if  a  solid,  Sec.    Q.  E.  D. 

PROP.  XXV.  THEOR. 

If  a  solid  parallelopiped  be  cut  by  a  plane  parallel  to 
two  of  its  opposite  planes,  it  divides  the  whole  into  two 
solids,  the  base  of  one  of  which  shall  be  to  the  base  of 
the  other,  as  the  one  solid  is  to  the  other.* 

Let  the  solid  parallelopiped  ABCD  be  cut  by  the  plane  EV, 
which  is  parallel  to  the  opposite  planes  AR,  HD;  and  divides  the 
whole  into  the  two  solids  ABFV,  EGCD;  as  the  base  AEFY  of 
the  first  is  to  the  base  EHCF  of  the  other;  so  is  the  solid  ABFV 
to  the  solid  EGCD. 

Produce  AH  both  ways,  and  take  any  number  of  straight  lines 
HM,  MN,  each  equal  to  EH,  and  any  number  AK,  KL  each 
equal  to  EA,  and  complete  the  parallelograms  LO,  KY,  HQ,  MS, 
and  the  solids  LP,  KR,  HU,  MT:  then  Ijecause  the  straight  lines 
LK,  KA,  AE  are  all  equal,  the  parallelograms  LO,  KY,  AF  are 


X 


B 


G 


i\P 


Z 


T 


V 


A  i^^        H     1 M        r^ 


^°-lX,H 


r 


\ 


O  Y  F        C        Q        S 

equal  (36.  1.):  and  likewise  the  parallelograms  KX,  BK,  AG,  (36. 
1.);  as  also  (24.  11.)  the  parallelograms  LZ,  KP,  AR,  because 
they  are  opposite  planes:  for  the  same  reason  the  parallelograms 
EC,  HQ,  MS  are  equal  (36.  1.);  and  the  parallelograms  HG,  HI, 
IN,  as  also  (24.  11.)  HD,  MU,  NT;  therefore  three  planes  of  the 
solid  LP,  are  equal  and  similar  to  three  planes  of  the  solid  KR,  as 
also  to  three  planes  of  the  solid  AV:  but  the  three  planes  oppo- 
site to  these  three  are  equal  and  similar  (24.  11.)  to  them  in  the 
several  solids,  and  none  of  their  solid  angles  are  contained  by  more 
than  three  plane  angles:  therefore  the  three  solids  LP,  KR,  AV 
are  equal  (C.  11.)  to  one  another:  for  the  same  reason,  the  three 
solids  ED,  HU,  MT  are  equal  to  one  another:  therefore  what 


See  Note. 


ROOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


181 


multiple  soever  the  base  LF  is  of  the  base  AF,  the  same  multiple 
is  the  solid  LV  of  the  solid  AV:  for  the  same  reason,  whatever 
multiple  the  base  NF  is  of  the  base  HF,  the  same  multiple  is  the 
solid  NV  of  the  solid  ED;  and  if  the  base  LF,  be  equal  to  the 
base  NF,  the  solid  LV  is  equal  (C.  11.)  to  the  solid  NV;  and  if 
the  base  LF  be  greater  than  the  base  NF,  the  solid  LV  is  greater 
than  the  solid  NV;  and  if  less,  less:  since  then  there  are  four 
magnitudes,  viz.  the  two  bases  AF,  FH,  and  the  two  solids  AV, 

X  B  G  I 


Xiv 

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.,     .TV- 

V 

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K 

A 

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E 

H 

M 

N 

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\rNi\, 

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O  Y  F        C         Q         S 

ED,  and  of  the  base  AF  and  solid  AV,  the  base  LF  and  solid  LV 
are  any  equimultiples  whatever;  and  of  the  base  FH  and  solid  ED, 
the  base  FN  and  solid  NV  are  any  equimultiples  whatever;  and  it 
has  been  proved,  that  if  the  base  LF  is  greater  than  the  base  FN, 
the  solid  LV  is  greater  than  the  solid  NV;  and  if  equal,  equal; 
and  if  less,  less.  Therefore  (5.  def.  5.)  as  the  base  AF  is  to  the 
base  FH,  so  is  the  solid  AV  to  the  solid  ED.  Wherefore,  if  a 
solid,  8cc.   Q.  E.  D. 

PROP.  XXVL  PROB. 

At  a  given  point  in  a  given  straight  line,  to  make  a 
solid  angle  equal  to  a  given  solid  angle  contained  by 
three  plane  angles.^ 

Let  AB  be  a  given  straight  line,  A  a  given  point  in  it,  and  D  a 
given  solid  angle  contained  by  the  three  plane  angles  EDC,  EDF, 
FDC:  it  is  required  to  make  at  the  point  A  in  the  straight  line 
AB  a  solid  angle  equal  to  the  solid  angle  D. 

In  the  straight  line  DF  take  any  point  F,  from  which  draw  (11. 
11.)  GF  perpendicular  to  the  plane  EDC,  meeting  that  plane  in 
G;  join  DG,  and  at  the  point  A  in  the  straight  line  AB  make  (23. 
1.)  the  angle  BAL  equal  to  the  angle  EDC,  and  in  the  plane  BAL 
make  the  angle  BAK  equal  to  the  angle  EDG:  {hen  make  AK 
equal  to  DG,  and  from  the  point  K  erect  (12.  11.)  KH  at  right 
angles  to  the  plane  BAL;  and  make  KH  equal  to  GF,  and  join 
AH:  then  the  solid  angle  at  A,  which  is  contained  by  the  three 
plane  angles  BAL,  BAH,  FL\L,  is  equal  to  the  solid  angle  at  D 
contained  by  the  three  plane  angles  EDC,  EDF,  FDC. 

Take  the  equal  straight  lines  AB,  DE,  and  join  HB,  KB,  FE, 
GE:  and  because  FG  is  perpendicular  to  the  plane  EDC,  it 
makes  right  angles  (3.  def.  11.)  with  every  straight  line  meeting 


*  See  Note. 


182  THE  ELEMENTS  OF  EUCLID.  BOOK  XI, 

it  in  that  plane:  therefore  each  of  the  angles  FGD,  FGE  is  a 
right  angle:  for  the  same  reason,  HKA,  HKB,  are  right  angles: 
and  because  KA,  AB  are  equal  to  GD,  DE,  each  to  each,  and 
contain  equal  angles,  therefore  the  base  BK  is  equal  (4.  1.)  to  the 
base  EG:  and  KH  is  equal  to  GF,  and  HKB,  FGE  are  right 
angles,  therefore  HB  is  equal  (4.  1.)  to  FE:  again,  because  AK, 
KH  are  equal  to  DG,  GF,  and  contain  right  angles,  the  base 
AH  is  equal  to  the  base  DF;  and  AB  is  equal  to  DE;  there- 
fore HA,  AB  are  equal  to  FD,  DE,  and  the  base  HB  is  equal 
to  the  base  FE,  therefore 
the  angle  BAH  is  equal 
(8.  1)  to  the  angle 
EDF:  for  the  same 
reason,   the  angle  HAL 

is    equal    to    the    angle  -o   /    /     \     \  j        E 
FDC.     Because    if    AL  ^    > 
and  DC  be  made  equal,         K'" 
and    KL,  HL,   GC,    EC  H  F 

be  joined,  since  the  whole  angle  GAL  is  equal  to  the  whole 
EDC,  and  the  parts  of  them  BAK,  EDG  are,  by  the  construction, 
equal:  therefore  the  remaining  angle  KAL  is  equal  to  the  re- 
maining angle  GDC:  and  because  KA,  AL  are  equal  to  GD,  DC, 
and  contain  equal  angles,  the  base  KL  is  equal  (4.  1.)  to  the  base 
GC:  and  KH  is  equal  to  GF,  so  that  LK,  KH  are  equal  to  CG, 
GF,  and  they  contain  right  angles:  therefore  the  base  HL  is 
equal  to  the  base  EC:  again,  because  HA,  AL  are  equal  to  FD, 
DC,  and  the  base  HL  to  the  base  EC,  the  angle  HAL  is  equal 
(8.  1.)  to  the  angle  FDC:  therefore,  because  the  three  plane  angles 
BAL,  BAH,  HAL,  which  contain  the  solid  angle  at  A,  are  equal 
to  the  three  plane  angles  EDC,  EDF,  FDC,  which  contain  the 
solid  angle  at  D,  each  to  each,  and  are  situated  in  the  same  order, 
the  solid  angle  at  A  is  equal  (B.  11.)  to  the  solid  angle  at  D. 
Therefore,  at  a  given  point  in  a  given  straight  line,  a  solid  angle 
has  been  made  equal  to  a  given  solid  angle  contained  by  three 
plane  angles.     Which  was  to  be  done. 


PROP.  XXVH.  PROB. 

To  describe  from  a  given  straight  line  a  solid  paral- 
lelopiped  similar  and  similarly  situated  to  one  given. 

Let  AB  be  the  given  straight  line,  and  CD  the  given  solid 
parallelopiped.  It  is  required  from  AB  to  describe  a  solid  pa- 
rallelopiped  similar  and  similarly  situated  to  CD. 

At  the  point  A  of  the  given  straight  line  AB,  make  (26.  11.) 
a  solid  angle  equal  to  the  solid  angle  at  C;  and  let  BAK,  KAH, 
HAB,  be  the  three  plane  angles  which  contain  it,  so  that  BAK 
be  equal  to  the  angle  ECG,  and  KAH  to  GCF,  and  HAB  to 
FCE:  and  as  EC  to  CG,  so  make  (12.  6.)  BA  to  AK:  and  as  GC 
to  CF,  so  make  (12.  6.)  KA  to  AHj  wherefore  ex  cequali  (22. 
5.)  as  EC  to  CF,  so  is  BA  to  AH^  complete  the  parallelogram 


BOOK   XI. 


THE  ELEMENTS  OF  EUCLID, 


183 


BH,  and  the  solid  AL: 
and  because,  as  EC  to 
CG,  so  BA  to  AK,  the 
sides  about  the  equal  an- 
gles ECG,  BAK  are 
proportionals:  therefore 
the  parallelogram  BK  is 
similar  to  EG.  For  the 
same  reason,  the  paral- 
lelogram KH  is  similar 
to  GF,  and  HB  to  FE 


\    H 


K 


T 


M 


D 


A  B 

Wherefore  three  parallelograms  of  the 
solid  AL  are  similar  to  three  of  the  solid  CD;  and  the  three  op- 
posite ones  in  each  solid  are  equal  (24.  11.)  and  similar  to  these, 
each  to  each.  Also,  because  the  plane  angles  which  contain  the 
solid  angles  of  the  figures  are  equal,  each  to  each,  and  situated 
in  the  same  order,  the  solid  angles  are  equal  (B.  1 1.),  each  to  each. 
Therefore  the  solid  AL  is  similar  (U.  def.  11.)  to  the  solid  CD. 
Wherefore  from  a  given  straight  line  AB,  a  solid  parallelopiped 
AL  has  been  described  similar  and  similarly  situated  to  the  given 
one  CD.     Which  v/as  to  be  done. 


PROP.  XXVIIL  THEOR. 

If  a  solid  parallelopiped  be  cut  by  a  plane  passing 
through  the  diagonals  of  two  of  the  opposite  planes;  it 
shall  be  cut  into  two  equal  parts.* 

Let  AB  be  a  solid  parallelopiped,  and  DE,  CF  the  diagonals  of 
the  opposite  parallelograms  AH,  GB,  viz.  those  which  are  drawn 
betwixt  the  equal  angles  in  each:  and  because  CD,  FE  are  each 
of  them  parallel  to  GA,  and  not  in  the  same  plane  with  it,  CD,FE 
are  parallel  (9.  11.);  wherefore  the  diagonals  CF,  DE  are  in  the 
plane  in  which  the  parallels  are,  and  are  ~ 

themselves  parallels  (16.  11.);  and  the 
plane  CDEF  shall  cut  the  solid  AB  into 
two  equal  parts. 

Because  the  triangle  CGF  is  equal  (.34. 
1.)  to  the  triangle  CBF,  and  the  triangle 
DAE,  to  DHE;  and  that  the  parallelo- 
gram CA  is  equal  (24.  11.)  and  similar 
to  the  opposite  one  BE;  and  the  parallel- 
ogram GE  to  CH:  therefore  the  prism 
contained  by  the  two  triangles  CGF,  DAE,  and  the  three  parallel- 
ograms CA,  GE,  EC,  is  equal  (C.  1 1.)  to  the  prism  contained  by 
the  two  triangles  CBF,  DHE,  and  the  three  parallelograms  BE, 
CH,  EC;  because  they  are  contained  by  the  same  number  of  equal 
and  similar  planes,  alike  situated,  and  none  of  their  solid  angles 
are  contained  by  more  than  three  plane  angles.  Therefore  the 
solid  AB  is  cut  into  two  equal  parts  by  the  plane  CDEF.  Q.  E.  D, 
'  N.  B.     The  insisting  straight  lines  of  a  parallelopiped,  men- 


''  See  Note. 


184 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI. 


tioned  in  the  next  and  some  following  propositions,  are  the  sides 
of  the  parallelograms  betwixt  the  base  and  the  opposite  plane  pa- 
rallel to  it.' 

PROP.  XXIX.  THEOR. 

Solid  parallelopipeds  upon  the  game  base,  and  of  the 
same  altitude,  the  insisting  Hues  of  which  are  termi- 
nated in  the  same  straight  Hues  in  the  plane  opposite  to 
the  base,  are  equal  to  one  another.^ 

Let  the  solid  parallelopipeds  AH,  AK  be  upon  the  same  base 
AB,  and  of  the  same  altitude,  and  let  their  insisting  straight  lines 
AF,  AG,  LM,  LN  be  terminated  in  the  same  straight  line  FN, 
and  CD,  CE,  BH,  BK  be  terminated  in  the  same  straight  line  DK; 
the  solid  AH  is  equal  to  the  solid  AK.f 

First,  let  the  parallelograms  DG,  HN,  which  are  opposite  to 
the  base  AB,  have  a  common  side  HG:  then,  because  the  solid  AH 
is  cut  by  the  plane  AGHC  passing  through  the  diagonals  AG, 
CH  of  the  opposite  planes  ALGF,  CBHD,  AH  is  cut  into  two 
equal  parts  (28.  11.)  by  the  plane  AGHC:  therefore  the  solid  AH 
is  double  of  the  prism  which  is  contained  betwixt  the  triangles 
ALG,  CBHj  for  the  same  reason         D  H  K 

because  the  solid  AK  is  cut  by 
the  plane  LGHB  through  the 
diagonals  LG,  BH  of  the  oppo- 
site planes  ALNG,  CBKH,  the 
solid  AK  is  double  of  the  same 
prism  which  is  contained  betwixt 
the  triangles  ALG,  CBH.  There- 
fore the  solid  AH  is  equal  to  the  solid  AK. 

But  let  the  parallelograms  DM,  EN  opposite  to  the  base  have 
no  common  side:  then,  because  CH,  CK  are  parallelograms,  CB  is 
equal  (34.  1.)  to  each  of  the  opposite  sides  DH,  EK^  wherefore 
DH  is  equal  to  EK:  add  or  take  away  the  common  part  HE;  then 
DE  is  equal  to  HK:  wherefore  also  the  triangle  CDE  is  equal 
(38.  1.)  to  the  triangle  BHK:  and  the  parallelogram  DC  is  equal 
(36.  1.)  to  the  parallelogram  HN:  for  the  same  reason  the  triangle 


D 


H 


E 


K 


D 


E 


H 


K 


\m7' 

I — J—/ — 


c 


N 


A  L  A  L 

AFG  is  equal  to  the  triangle  LMN,  and  the  parallelogram  CF 
is  equal  (24.  11.)  to  the  parallelogram  ^BN,  and  CG  to  BNj  for 
they  are  opposite.     Therefore  the  prism  which  is  contained  by 


See  Note. 


See  the  figures. 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


185 


the  two  triangles  AFG,  CDE,  and  the  three  parallelograms  AD, 
DG,  GC,  is  equal  (C.  11.)  to  the  prism  contained  by  the  two  tri- 
angles LMN,  BHK,  and  the  three  parallelograms  BM,  MK,  KL. 
If  therefore  the  prism  LMNBHK  be  taken  from  the  solid  of  which 
the  base  is  the  parallelogram  AB,  and  in  which  FDKN  is  the  one 
opposite  to  it;  and  if  from  this  same  solid  there  be  taken  the  prism 
AFGCDE,  the  remaining  solid,  viz.  the  parallelopiped  AH,  is 
equal  to  the  remaining  parallelopiped  AK.  Therefore,  solid  pa- 
rallelopipeds,  Sec.  Q.  E.  D. 

PROP.  XXX.  THEOR. 

Solid  parallelepipeds  upon  the  same  base,  and  of  the 
same  altitude,  the  insisting  straight  lines  of  which  are 
not  terminated  in  the  same  straight  lines  in  the  plane 
opposite  to  the  base,  are  equal  to  one  another.* 

Let  the  parallelopipeds  CM,  CN  be  upon  the  same  base  AB, 
and  of  the  same  altitude,  but  their  insisting  straight  lines  AF, 
AG,  LM,  LN,  CD,  CE,  BH,  BK,  not  terminated  in  the  same 
straight  lines;  the  solids  CM,  CN,  are  equal  to  one  another. 

Produce  FD,  MH,  and  NG,  KE;  and  let  them  meet  one  another 
in  the  points  O,  P,  Q,  R;  and  join  AC),  LP,  BQ,  CR:  and  because 
the  plane  LBHN  is  parallel  to  the  opposite  plane  ACDF,  and  that 

N  K 


the  plane  LBHM  is  that  in  which  are  the  parallels  LB,  MHPQ, 
in  which  also  is  the  figure  BLPQ,  and  the  plane  ACDF  is  that  in 
which  are  the  parallels  AC,  FDOR,  in  which  also  is  the  figure 
CAOR;  therefore  the  figures  BLPQ,  CAOR  are  in  parallel  planes; 
in  like  manner,  because  the  plane  ALNG  is  parallel  to  the  oppo- 
site plane  CBKE,  and  that  the  plane  ALNG  is  that  in  which  are 
the  parallels  AL,  OPGN,  in  which  also  is  the  figure  ALPO;  and 
the  plane  CBKE  is  that  in  which  are  the  parallels  CB,  RQEK,  in 
which  also  is  the  figure  CBQR;  therefore  the  figures  ALPO, 
CBQR  are  in  parallel  planes;  and  the  planes  ACBL,  ORQP  are 


566 


Note. 


2  A 


186 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI. 


parallel;  therefore  the  solid  CP  is  a  parallelopiped;  but  the  solid 
CM,  of  which  the  base  is  ACBL,  to  which  FDHM  is  the  opposite 
parallelogram,  is  equal  (29.  1 1 .)  to  the  solid  CP,  of  which  the  base 

N  K 


is  the  parallelogram  ACBL,  to  which  ORQP  is  the  one  opposite; 
because  they  are  upon  the  same  base,  and  their  insisting  straight 
lines  AF,  AO,  CD,  CR;  LM,  LP,  BH,  BQ  are  in  the  same  straight 
lines  FR,  MQ:  and  the  solid  CP  is  equal  (29.  11.)  to  the  solid  CN: 
for  they  are  upon  the  same  base  ACBL,  and  their  insisting  straight 
lines  AO,  AG,  LP,  LN;  CR,  CE,  BQ,  BK,  are  in  the  same  straight 
lines  ON,  RK:  therefore  the  solid  CM  is  equal  to  the  solid  CN. 
Wherefore,  solid  parallelopipeds,  &;c.  Q.  E.  D.. 


PROP.    XXXL  THEOR. 

Solid  parallelopipeds  which  are  upon  equal  bases 
and  of  the  same  altitude,  are  equal  to  one  another.*" 

Let  the  solid  parallelopipeds  AE,  CF  be  upon  equal  bases  AB, 
CD,  and  be  of  the  same  altitude;  the  solid  AE  is  equal  to  the  solid 
CF. 

First,  let  the  insisting  straight  lines  be  at  right  angles  to  the 
bases  AB,  CD,  and  let  the  bases  be  placed  in  the  same  plane,  and 
so  as  that  the  sides  CL,  LB  be  in  a  straight  line;  therefore  the 
straight  line  LM  which  is  at  right  angles  to  the  plane  in  which 
the  bases  are,  in  the  point  L,  is  common  (13.  11.)  to  the  two  solids 
AE,  CF;  let  the  other  insisting  lines  of  the  solids  be  AG,  HK, 
BE;  DF,  OP,  CN:  and  first,  let  the  angle  ALB  be  equal  to  the 
angle  CLD;  then  AL,  LD  are  in  a  straight  line  (14.  1.).  Produce 
OD,  HB,  and  let  them  meet  in  Q,  and  complete  the  solid  paral- 
lelopiped LR,  the  base  of  which  is  the  parallelogram  LQ,  and  of 
which  LM  is  one  of  its  insisting  straight  lines:  therefore,  because 
the  parallelogram  AB  is  equal  to  CD;  as  the  base  AB  is  to  the 
base  LQ,  so  is  (7.  5.)  the  base  CD  to  the  same  LQ:  and  because 
the  solid  parallelopiped  AR  is  cut  by  the  plane  LMEB,  which  is 

**  See  Note. 


BOOK  XI. 


THE  ELEMENTS  OF    EUCLID. 


187 


parallel  to  the  opposite  planes  AK,  DR;  as  the  base  AB  is  to  the 
base  LQ,  so  is  (25.  11.)  the  solid  AE  to  the  solid  LR:  for  the  same 
reason,  because  the  solid  parallelopiped  CR  is  cut  by  the  plane 
LMFD,  which  is  parallel  to  the  opposite  planes  CP,  BRj  as  the 


base  CD  to  the  base 
LQ,  so  is  the  solid 
CF  to  the  solid  LR: 
but  as  the  base  AB 
to  the  base  LQ,  so 
the  base  CD  to  the 
base  LQ,  as  before 
was  proved:  there- 
fore as  the  solid  AE 
to  the  solid  LR,  so 
is  the  solid  CF  to  the 


R 


O 


\N 

X.LM 

\^1 

-p 

L-j 
N-^, 

D 

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^ 

G 

\ 

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K 

\ 

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^ 

^ 

L 

"^ 

^^ 

S 

-^.^^ 

A    S 

therefore  the 


H     T 
solid  AE  is 


solid  LR;  and 
equal  (9.  5.)  to  the  solid  CF. 

But  let  the  solid  parallelepipeds  SE,  CF  be  upon  equal  bases 
SB,  CD,  and  be  of  the  same  altitude,  and  let  their  insisting  straight 
lines  be  at  right  angles  to  the  bases;  and  place  the  bases  SB,  CD 
in  the  same  plane,  so  that  CL,  LB  be  in  a  straight  line;  and  let 
the  angles  SLB,  CLD  be  unequal;  the  solid  SE  is  also  in  this  case 
equal  to  the  solid  CF:  produce  DL,  TS,  until  they  meet  in  A,  and 
from  B  draw  BH  parallel  to  DA;  and  letHB,  OD  produced  meet 
in  Q,  and  complete  the  solids  AE,  LR;  therefore  the  solid  AE,  of 
which  the  base  is  the  parallelogram  LE,  and  AK  the  one  opposite 
to  it,  is  equal  (29.  11.)  to  the  solid  SE,  of  which  the  base  is  LE, 
and  to  which  SX  is  opposite:  for  they  are  upon  the  sanae  base  LE, 
and  of  the  same  altitude,  and  their  insisting  straight  lines,  viz. 
LA,  LS,  BH,  BT;  MG,  MV,  EK,  EX,  are  in  the  same  straight 
lines  AT,  GX;  and  because  the  parallelogram  AB  is  equal  (35. 
1.)  to  SB,  for  they  are  upon  the  same  base  LB,  and  between  the 
same  parallels  LB,  AT;  and  that  the  base  SB  is  equal  to  the  base 
CD;  therefore  the  base  AB  is  equal  to  the  base  CD,  and  the  angle 
ALB  is  equal  to  the  P  F  R 

angle   CLD;   there- 
fore, by  the  first  case, 

the  solid  AE  is  equal  ^!!>v>>ri I  ^^Np^  X 

to  the  solid  CF;  but 
the  solid  AE  is  equal  O 
to  the  solid  SE,  as 
was  demonstrated ; 
therefore  the  solid 
SE  is  equal  to  the 
solid  CF. 

But  if  the  insisting  straight  lines  AG,  HK,  BE,  LM;  CN,  RS, 
DF,  OP,  be  not  at  right  angles  to  the  bases  AB,  CD;  in  this  case 
likewise  the  solid  AE,  is  equal  to  the  solid  CF:  from  the  points 
G,  K,  E,  M;  N,  S,  F,  P,  draw  the  straight  lines  GQ,  KT,  EV, 
MX;  NY,  SZ,  FI,  PU,  perpendicular  (ll.  II.)  to  the  plane  in 
which  are  the  bases  AB,  CD;  and  let  them  meet  it  in  the  points 
Q,  T,  V,  X;  Y,  Z,  I,  U  and  join  QT,  TV,  VX,  XQ;  YZ,  ZI,  lU, 
UY:  then  because  GQ,  KT  are  at  right  angles  to  the  same  plane, 


N^N 

\A1 

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E 

D 

^ 

■^^O' 

^ 

p-^ 

G 

k 

B 

K 

\ 

\ 

C 

L 

^ 

r-^ 

S 

A     S 


H     T 


188 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI. 


M 

E 

yy*  ^ 

>< 

GJ 

\ 

/^  V^' 

y 

wy^ 

\>c 

^ 

^ 

\ 

F 


^ 


AHQT  C  R  Y  Z 

they  are  parallel  (6.  1 1.)  to  one  another:  and  MG,  EK  are  paral- 
lels; therefore  the  plane  MQ,  ET,  of  which  one  passes  through 
MG,  GQ,  and  the  other  through  EK,  KT,  which  are  parallel  to 
MG,  GQ,  and  not  in  the  same  plane  with  them,  are  parallel  (15. 
1 1.)  to  one  another.  For  the  same  reason  the  planes  MV,  GT  are 
parallel  to  one  another:  therefore  the  solid  QE  is  a  parallelopiped: 
in  like  maimer  it  may  be  proved,  that  the  solid  YF  is  a  parallelo- 
piped: but,  from  what  has  been  demonstrated,  the  solid  EQ  is 
equal  to  the  solid  FY,  because  they  are  upon  equal  bases  MK,  PS, 
and  of  the  same  altitude,  and  have  their  insisting  straight  lines  at 
right  angles  to  the  bases:  and  the  solid  EQ  is  equal  (29.  or  30.  1 1.) 
to  the  solid  AEj  and  the  solid  FY  to  the  solid  CF;  because  they 
are  upon  the  same  bases  and  of  the  same  altitude:  therefore  the 
solid  AE  is  equal  to  the  solid  CF.  Wherefore  solid  parallelo- 
pipeds,  &:c.  Q.  E.  D. 

PROP.  XXXII.  THEOR. 

Solid  parallelepipeds  which  have  the  same  altitude, 
are  to  one  another  as  their  bases.^ 

Let  AB,  CD  be  solid  parallelepipeds  of  the  same  altitude;  they 
are  to  one  another  as  their  bases;  that  is,  as  the  base  AE  to  the 
base  CF,  so  is  the  solid  AB  to  the  solid  CD. 

To  the  straight  line  FG  apply  the  parallelogram  FH  equal  (Cor. 
45.  1.)  to  AE,  so  that  the  angle  FGH  be  equal  to  the  angle  LCG, 
and  complete  the  solid  parallelopiped  GK  upon  the  base  FH,  one 
of  whose  insisting  lines  is  FD,  whereby  the  solids  CD,  GK  must 
be  of  the  same  altitude;  therefore  the  solid  AB  is  equal  (31.  11.) 


B 


D 


to  the  solid  GK, 
because  they  are 
upon  equal  bases 
AE,  FH,  and  are 
of  the  same  alti- 
tude: and  because 
the  solid  parallelo- 
piped   CK  is    cut 

by  the  plane  DG  A  M  C  G  H 

which  is  parallel  to  its  opposite  planes,  the  base  HF  is  (25.  1 1.)  to 
the  base  FC,  as  the  solid  WVi  to  the  solid  DC:  but  the  base  HF 
IS  equal  to  the  ba5e  AE,  and  the  solid  GK  to  the  solid  AB:  there- 
fore, as  the  base  AE  to  the  base  CF,  so  is  the  solid  AB  to  the 
solid  CD.     Wherefore  solid  paralleloplpeds,  8cc.  Q.  E.  D. 


*^  See  Note. 


BOOK  XI. 


THE  ELKMENTS  OF  EUCLID. 


189 


Cor.  From  this  it  is  manifest  that  prisms  upon  triangular  bases, 
of  the  same  altitude,  are  to  one  another  as  their  bases. 

Let  the  prisms,  the  bases  of  which  are  the  triangles  AEM,  CFG, 
and  NBO,  PDQ  the  triangles  opposite  to  them,  have  the  same  al- 
titude; and  complete  the  parallelograms  AE,  CF,  and  the  solid 
parallelopipeds  AB,  CD,  in  the  first  of  which  let  MO,  and  in  the 
other  let  GQ  be  one  of  the  insisting  lines.  And  because  the  solid 
parallelopipeds  AB,  CD  have  the  same  altitude,  they  are  to  one 
another  as  the  base  AE  is  to  the  base  CF;  wherefore  the  prisms, 
which  are  their  halves  (28.  1 1.)  are  to  one  another  as  the  base  AE 
to  the  base  CF;  that  is,  as  the  triangle  AEM  to  the  triangle  CFG. 

PROP.  XXXIII.  THEOR. 
Similar  solid  parallelopipeds  are  one  to  another  in 
the  triplicate  ratio  of  their  homologous  sides. 

Let  AB,  CD  be  similar  solid  parallelopipeds,  and  the  side  AE 
homologous  to  the  side  CF:  the  solid  AB  has  to  the  solid  CD  the 
triplicate  ratio  of  that  which  AE  has  to  CF. 

Produce  AE,  GE,  HE,  and  in  these  produced  take  EK  equal  to 
CF,  EL  equal  to  FN,  and  EM  equal  to  FR;  and  complete  the  pa- 
rallelogram KL,  and  the  solid  KO:  because  KE,  EL  are  equal  to 
CF,  FN,  and  the  angle  KEL,  equal  to  the  angle  CFN,  because  it 
is  equal  to  the  angle  AEG,  which  is  equal  to  CFN,  by  reason  that 
the  solids  AB,  CD  are  similar;  therefore  the  parallelogram  KL  is 
similar  and  equal  to  the  parallelogram  CN:  for  the  same  reason, 
the  parallelogram  MK  is  similar  and  equal  to  CR,  and  also  OE  to 
FD.  Therefore  three  parallelograms  of  the  solid  KO  are  equal 
and  similar  to  three  B  X 

parallelograms  of  the  -j 

solid   CD ;    and    the     , 

three   opposite    ones      \ 
in  each  solid  are  equal 
(24.  11.)  and  similar  ^ 

to  these:  therefore  the 
solid  KO  is  equal  (C. 
1 1 .)  and  similar  to  the 
solid  CD :  complete 
the  parallelogram  GK 
and  complete  the  so- 
lids EX,  LP  upon  the  O 
bases  GK,  KL,  so  that  EH  be  an  insisting  straight  line  in  each  of 
them,  whereby  they  must  be  of  the  same  altitude  with  the  solid 
AB:  and  because  the  solids  AB,  CD  are  similar,  and,  by  permu- 
tation, as  AE  is  to  CF,  so  is  EG  to  FN,  and  so  is  EH  to  FR;  and 
FC  is  equal  to  EK,  and  FN  to  EL,  and  FR  to  EM:  therefore  as 
AE  to  EK,  so  is  EG  to  EL,  and  so  is  HE  to  EM:  but,  as  AE  to 
EK,  so  (1.  6.)  is  the  parallelogram  AG  to  the  parallelogram  GK: 
and  as  GE  to  EL,  so  is  (I.  e!)  GK  to  KL,  and  as  HE  to  EM,  so 
(1.6.)  is  PE  to  KM:  therefore  as  the  parallelogram  AG  to  the 
parallelogram  GK,  so  is  GK  to  KL,  and  PE  to  KM:  but  as  AG  to 
GK,  so  (25.  11.)  is  the  solid  AB  to  the  solid  EX;  and  as  GK  to 


Xrz^J 


R 


\ 

\H\ 

P 

G 

\ 

\ 

1 

\ 

,      \ 

\ 

E 

J 

\               V 

\ 

\ 

i\  A 

1. 

M 

M 

\ 

J  90  THE  ELEMENTS  OF  EUCLID.  BOOK  XI, 

KL,  SO  (25.  11.)  is  the  solid  EX  to  the  solid  PL;  and  as  PE  to 
KM,  so  (25.  11.)  is  the  solid  PL  to  the  solid  KO:  and  therefore 
as  the  solid  AB  to  the  solid  EX,  so  is  EX  to  PL,  and  PL  to  KO: 
but  if  four  magnitudes  be  continual  proportionals,  the  first  is  said 
to  have  to  the  fourth,  the  triplicate  ratio  of  that  which  it  has  to 
the  second:  therefore  the  solid  AB  has  to  the  solid  KO  the  tri- 
plicate ratio  of  that  which  AB  has  to  EX:  but  as  AB  is  to  EX, 
so  is  the  parallelogram  AG  to  the  parallelogram  GK,  and  the 
straight  line  AE  to  the  straight  line  EK.  Wherefore  the  solid 
AB  has  to  the  solid  KO  the  triplicate  ratio  of  that  which  AE 
has  to  EK.  And  the  solid  KO  is  equal  to  the  solid  CD,  and  the 
straight  line  EK  is  equal  to  the  straight  line  CF.  Therefore  the 
solid  AB  has  to  the  solid  CD  the  triplicate  ratio  of  that  which 
the  side  AE  has  to  the  homologous  side  CF,  Sec.  Q.  E.  D. 

Cor.  From  this  it  is  manifest,  that,  if  four  straight  lines  be 
continual  proportionals,  as  the  first  is  to  the  fourth,  so  is  the  solid 
parallelopiped  described  from  the  first  to  the  similar  solid  si- 
milarly described  from  the  second 5  because  the  first  straight  line 
has  to  the  fourth  the  triplicate  ratio  of  that  which  it  has  to  the 
second. 

PROP.  D.  THEOR. 

Solid  parallelepipeds  contained  by  parallelograms 
equiangular  to  one  another,  each  to  each,  that  is,  of 
which  the  solid  angles  are  equal,  each  to  each,  have  to 
one  another  the  ratio  which  is  the  same  with  the 
ratio  compounded  of  the  ratios  of  their  sides.* 

Let  AB,  CD  be  solid  parallelopipeds,  of  which  AB  is  contain- 
ed by  the  parallelograms  AE,  AF,  AG,  equiangular,  each  to  each, 
to  the  parallelograms  CH,  CK,  CL,  which  contains  the  solid  CD. 
The  ratio  which  the  solid  AB  has  to  the  solid  CD,  is  the  same 
with  that  which  is  compounded  of  the  ratios  of  the  sides  AM  to 
DL,  AN  to  DK,  and  AO  to  DH. 

Produce  MA,  NA,  OA,  to  P,  Q,  R,  so  that  AP  be  equal  to 
DL,  AQ  to  DK,  and  AR  to  DH;  and  complete  the  solid  paral- 
lelopiped AX  contained  by  the  parallelograms  AS,  AF,  AV, 
similar  and  equal  to  CH,  CK,  CL,  each  to  each.  Therefore  the 
solid  AX  is  equal  (C.  11.)  to  the  solid  CD.  Complete  like- 
wise the  solid  AY,  the  base  of  which  is  AS,  and  of  which  AO  is 
one  of  its  insisting  straight  lines.  Take  any  straight  line  a,  and 
as  MA  to  AP,  so  make  a  to  b,  and  as  NA  to  AQ,  so  make  b  to 
cj  and  as  AO  to  AR,  so  c  to  d:  then  because  the  parallelogram 
AE  is  equiangular  to  AS,  AE  is  to  AS,  as  the  straight  line  a' to 
c,  as  is  demonstrated  in  the  23d  prop,  book  6:  and  the  solids 
AB,  AY,  being  betwixt  the  parallel  planes  BOY,  EAS,  are  of  the 
same  altitude  Therefore  the  solid  AB  is  to  the  solid  AY,  as 
(32.  11.)  the  base  AE  to  the  base  AS;  that  is,  as  the  straight 
line  a  is  to  c.  And  the  solid  AY,  is  to  the  solid  AX,  as  (25. 
11.)  the  base  OQ   is  to  the  base  QR^   that  is,  as  the   straight 

*  See  Note. 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 

B  G 


191 


D 

L 

\ 

N. 

R 

LI 

\ 

\ 

i: 


c 


a- 


Cm' 

d 


XF 

\^ 

J 

N 

\ 

r 

\ 

\ 

\ 

M 

A 

Q 

Yi 

T 

\ 

\ 

s 


V  X 

line  OA  to  AR;  that  is,  as  the  straight  line  c  to  the  straight  line 
d.  And  because  the  solid  AB  is  to  the  solid  AY,  as  a  is  to  c, 
and  the  solid  AY  to  the  solid  AX  as  c  is  to  dj  ex  sequali^  the 
solid  AB  is  to  the  solid  AX,  or  CD  which  is  equal  to  it,  as  the 
straight  line  a  is  to  d.  But  the  ratio  of  a  to  d  is  said  to  be  com- 
pounded (def.  A.  5.)  of  the  ratios  of  a  to  b,  b  to  c,  and  c  to  d, 
which  are  the  same  with  ratios  of  the  sides  MA,  to  AP,  NA  to 
AQ,  and  OA  to  AR,  each  to  each.  And  the  sides  AP,  AQ,  AR 
are  equal  to  the  sides  DL,  DK,  DH,  each  to  each.  Therefore 
the  solid  AB  has  to  the  solid  CD  the  ratio  which  is  the  same 
with  that  which  is  compounded  of  the  ratios  of  the  sides  AM  to 
DL,  AN  to  DK,  and  AO  to  DH.     Q.  E.  D. 

PROP.  XXXIV.  THEOR. 

The  bases  and  altitudes  of  equal  solid  parallelopipeds, 
are  reciprocally  proportional:  and  if  the  bases  and  al- 
titudes be  reciprocally  proportional,  the  solid  parallelo- 
pipeds are  equal."* 

Let  AB,  CD  be  equal  solid  parallelopipeds;  their  bases  are  re- 
ciprocally proportional  to  their  altitudes;  that  is,  as  the  base  EH 
is  to  the  base  NP,  so  is  the  altitude  of  the  solid  CD  to  the  alti- 
tude of  the  solid  AB. 

First,  let  the  insisting  straight  lines,  AG,  EF,  LB,  HK;  CM, 
NX,  OD,  PR  be  at  right  angles  to  the  bases.  As  the  base  EH  to 
the  base  NP,  so  is  CM  to  AG. 
If  the  base  EH  be  equal  to 
the  base  NP,  then  because 
the  solid  AB  is  likewise  equal 
to  the  solid  CD,  CM  shall  be 
equal  to  AG.  Because  if 
the  bases  EH,  NP  be  equal, 
but  the  altitudes  AG,  CM 
be  not   equal,  neither    shall 


N 


A  E  C 

the  solid  AB  be  equal  to  the  solid  CD.  But  the  solids  are  equal, 
by  the  hypothesis.  Therefore  the  altitude  CM  is  not  unequal  to 
the  altitude  AG;  that  is,  they  are  equal.  Wherefore,  as  the  base 
EH  to  the  base  NP,  so  is  CM  to  AG. 


*  See  Note. 


192 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI. 


R 


D 


K 


13 


\^ 

G 

L 

\ 

\ 

Y 


\ 

\'i 

V 

T 

\ 

o 

\ 

\ 

X 


Next,  let  the  bases  EH,  NP  not  be  equal;  but  EH  greater  than 
the  other:  since  then  the  solid  AB  is  equal  to  the  solid  CD,  CM, 
is  therefore  greater  than 
AG:  for  if  it  be  not,  nei- 
ther also,  in  this  case, 
would  the  solids  AB,  CD 
be  equal,  which,  by  the  hy- 
pothesis, are  equal.  Make 
then  CT  equal  to  AG,  and 
complete  the  solid  paral- 
lelepiped CV  of  which  H 
the  base  is  NP,  and  alti- 
tude CT.  Because  the  so-  A  E  C  N 
lid  AB  is  equal  to  the  solid  CD,  therefore  the  solid  AB  is  to  the 
solid  CV,  as  (7.  5.)  the  solid  CD  to  the  solid  CV.  But  as  the 
solid  AB  to  the  solid  CV,  so  (32.  1 1.)  is  the  base  EH  to  the  base 
NP;  for  the  solids  AB,  CV  are  of  the  same  altitude;  and  as  the 
solid  CD  to  CV,  so  (25.  1 1.)  is  the  base  MP  to  the  base  PT,  and 
so  (1.  6.)  is  the  straight  line  MC  to  CT;  and  CT  is  equal  to  AG. 
Therefore,  as  the  base  EH  to  the  base  NP,  so  is  MC  to  AG. 
Wherefore,  the  bases  of  the  solid  parallelopipcds  AB,  CD  are  re- 
ciprocally proportional  to  their  altitudes. 

Let  now  the  bases  of  the  solid  parallelepipeds  AB,  CD  be  re- 
ciprocally proportional  to  their  altitudes;  viz.  as  the  base  EH  to 


K 


B 


R 


D 


H 


the  base  NP,  so  the  altitude 

of  the  solid  CD  to  the  alti- 
tude of  the  solid  AB;   the 

solid  AB  is  equal  to  the  solid 

CD.     Let  the  insisting  lines 

be,  as  before,  at  right  angles 

to  the  bases.     Then,  if  the 

base  EH  be  equal  to  the  base 

NP,  since  EH  is  to  NP,  as         A  E  C  N 

the  altitude  of  the  solid  CD  is  to  the  altitude  of  the  solid  AB, 
therefore  the  altitude  of  CD  is  equal  (A.  5.)  to  the  altitude  of  AB. 
But  solid  parallelepipeds  upon  equal  bases,  and  of  the  same  alti- 
tude, are  equal  (31.  11.)  to  one  another:  therefore  the  solid  AB  is 
equal  to  the  solid  CD. 

But  let  the  bases  EH,  NP  be  unequal,  and  let  EH  be  the  greater 
of  the  two.     Therefore,  since  as  the  base  EH  to  the  base  NP,  so 


is  CM  the  altitude  of  the 
solid  CD  to  AG  the  alti- 
tude of  AB,  CM  is  great- 
er (A.  5.)  than  AG.  Again, 
take  CT  equal  to  AG,  and 
complete,  as  before,  the 
solid  CV.  And  because 
the  base  EH  is  to  the  base 
NP,  as  CM  to  AG,  and 
that  AG  is  equal  to  CT, 
therefore  the  base  EH  is 
to  the  base  NP,  as  MC  to 


R 


D 


K 


B 


K 


G 


\ 


L 


F 


M 


V 


S^ 


^ 


E 


N 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


193 


CT.  But  as  the  base  EH  is  to  NP,  so  (32.  1 1.)  is  the  solid  AB 
to  the  solid  CV;  for  the  solids  AB,  CV  are  of  the  same  altitude^ 
and  as  MC  to  CT,  so  is  the  base  MP  to  the  base  PT,  and  the 
solid  GD  to  the  solid  (25.  11.)  CV:  and  therefore  as  the  solid  AB 
to  the  solid  CV,  so  is  the  solid  CD  to  the  solid  CV;  that  is,  each 
of  the  solids  AB,  CD  has  the  same  ratio  to  the  solid  CVj  and 
therefore  the  solid  AB  is  equal  to  the  solid  CD. 

Second  general  case.  Let  the  insisting-  straight  lines  FE,  BL, 
GA,  KHj  XN,  DO,  MC,  RP  not  be  at  right  angles  tp^the  bases 
of  the  solidsj  and  from  the  points  F,  B,  K,  G^  X,  D,"^,  M* draw- 
perpendiculars  to  the  planes  in  which  are  the  bases  EH,  Nit*,  meet- 
ing those  planes  in  the  points  S,  Y,  V,  T;  Q,  I,  U,  Z;  and  com- 
plete the  solids  FV,  XU,  which  are  parallelopipeds,  as  was  proved 
in  the  last  part  of  Prop.  31.  of  this  Book.  In  this  case  likewise, 
if  the  solids  AB,  CD  be  equal,  their  bases  are  reciprocally  propor- 
tional to  their  altitudes,  viz.  the  base  EH  to  the  base  NP,  as  the 
altitude  of  the  solid  CD  to  the  altitude  of  the  solid  AB.  Because 
the  solid  AB  is  equal  to  the  solid  CD,  and  that  the  solid  BT  is 
equal  (29.  or  30.  11.)  to  the  solid  BA,  for  they  are  upon  the  same 
base  FK,  and  of  the  same  altitude;  and  that  the  solid  DC  is  equal 


K 


B 


R 


D 


A  E 

(29.  or  30.  11.)  to  the  solid  DZ,  being  upon  the  same  base  XR, 
and  of  the  same  altitude;  therefore  the  solid  BT  is  equal  to  the 
solid  DZ:  but  the  bases  are  reciprocally  proportional  to  the  alti- 
tudes of  equal  solid  parallelopipeds  of  which  the  insisting  straight 
lines  are  at  right  angles  to  their  bases,  as  before  was  proved. 
Therefore  as  the  base  FK  to  the  base  XR,  so  is  the  altitude  of  the 
solid  DZ  to  the  altitude  of  the  solid  BT:  and  the  base  FK  is  equal 
to  the  base  EH,  and  the  base  XR  to  the  base  NP.  Wherefore, 
as  the  base  EH  to  the  base  NP,  so  is  the  altitude  of  the  solid  DZ 
to  the  altitude  of  the  solid  BT:  but  the  altitudes  of  the  solids  DZ, 
DC,  as  also  of  the  solids  BT,  BA  are  the  same.  Therefore  as  the 
base  EH  to  the  base  NP,  so  is  the  altitude  of  the  solid  CD  to  the 
altitude  of  the  solid  AB;  that  is,  the  bases  of  the  solid  parallelo- 
pipeds AB,  CD  are  reciprocally  proportional  to  their  altitudes. 

Next,  let  the  bases  of  the  solids  AB,  CD  be  reciprocally  pro- 
portional to  their  altitudes,  viz.  the  base  EH  to  the  base  NP,  as 
the  altitude  of  the  solid  CD  to  the  altitude  of  the  solid  AB;  the 
solid  AB  is  equal  to  the  solid  CD:  the  same  construction  being 
2.  B 


194 


% 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI« 


made:  because  as  the  base  EH  to  the  base  NP,  so  is  the  altitude 
of  the  solid  CD  to  the  altitude  of  the  solid  AB^  and  that  the  base 
EH  is  equal  to  the  base  FK^  and  NP  to  XR;  therefore  the  base 
FK  is  to  the  base  XR,  as  the  altitude  of  the  solid  CD  to  the  alti- 
tude of  AB.     But  the  altitudes  of  the  solids  AB,  BT  are  the  same, 


K 


B 


R 


D 


^X 


E 


N 


as  also  of  CD  and  DZ;  therefore  as  the  base  FK  to  the  base  XR, 
so  is  the  altitude  of  the  solid  DZ  to  the  altitude  of  the  solid  BT: 
wherefore  the  bases  of  the  solids  BT,  DZ  are  reciprocally  propor- 
tional to  their  altitudes^  and  their  insisting  straight  lines  are  at 
right  angles  to  the  basesj  wherefore,  as  was  before  proved,  the 
solid  BT  is  equal  to  the  solid  DZ:  but  BT  is  equal  (29.  or  30.  11.) 
to  the  solid  BA,  and  DZ  to  the  solid  DC,  because  they  are  upon 
the  same  bases,  and  of  the  same  altitude.  Therefore  the  solid 
AB  is  equal  to  the  solid  CD.     Q.  E.  D. 

PROP.  XXXV.  THEOR. 

If,  from  the  vertices  of  two  equal  plane  angles,  there 
be  drawn  two  straight  lines  elevated  above  the  planes 
in  which  the  angles  are,  and  containing  equal  angles 
with  the  sides  of  those  angles,  each  to  each;  and  if  in 
the  lines  above  the  planes  there  be  taken  any  points, 
and  from  them  perpendiculars  be  drawn  to  the  planes 
in  which  the  first  named  angles  are :  And  from  the  points 
in  which  they  meet  the  planes,  straight  lines  be  drawn 
to  the  vertices  of  the  angles  first  named ;  these  straight 
lines  shall  contain  equal  angles  with  the  straight  lines 
which  are  above  the  planes  of  the  angles.* 

Let  BAC,  EDF  be  two  ec|ual  plane  angles;  and  from  the  points 
A,  D  let  the  straight  lines  AG,  DM  be  elevated  above  the  planes 
of  the  angles,  making  equal  angles  with  their  sides,  each  to  each, 
viz.  the  angle  GAB  equal  to  the  angle  MDE,  and  GAC  to  MDF: 
and  in  AG,  DM  let  any  poiuis  (i,  M  be  taken,  and  from  them  let 


*  Sec  Note. 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


195 


perpendiculars  GL,  MN  be  drawn  to  the  planes  BAG,  EDF,  meet- 
ing these  planes  in  the  points  L,  N,  and  join  LA,  ND:  the  angle 
GAL  is  equal  to  the  angle  MDN. 

Make  AH  equal  to  DM,  and  through  H  draw  HK  parallel  to 
GL.  But  GL  is  perpendicular  to  the  plane  BACj  wherefore  HK 
is  perpendicular  (8.  11.)  to  the  same  plane:  from  the  points  K,  N 
to  the  straight  lines  AB,  AC,  DE,  DF,  draw  perpendiculars  KB, 

D 

.A. 


KG,  NE,  NF^  and  join  HB,  BG,  ME,  EF:  Because  HK  is  perpen- 
dicular to  the  plane  BAG,  the  plane  HBK  which  passes  through 
HK  is  at  right  angles  (18.  11.)  to  the  plane  BAG:  and  AB  is 
drawn  in  the  plane  BAG  at  right  angles  to  the  common  section 
BK  of  the  two  planes;  therefore  AB  is  perpendicular  (4.  def.  1 1.) 
to  the  plane  HBK,  and  makes  right  angles  (3.  def.  1 1.)  with  every 
straight  line  meeting  it  in  that  plane.  But  BH  meets  it  in  that 
planej  therefore  ABH  is  a  right  angle.  For  the  same  reason, 
DEM  is  a  right  angle,  and  is  therefore  equal  to  the  angle  ABH: 
and  the  angle  HAB  is  equal  to  the  angle  MDE.  Therefore  in  the 
two  triangles  HAB,  MDE  there  are  two  angles  in  one  equal  to 
two  angles  in  the  other,  each  to  each,  and  one  side  equal  to  one 
side,  opposite  to  one  of  the  equal  angles  in  each,  viz.  HA  equal  to 
DM;  therefore  the  remaining  sides  are  equal  (26.  1.)  each  to  each: 
wherefore  AB  is  equal  to  DE.  In  the  same  manner,  if  HG  and 
MF  be  joined,  it  may  be  demonstrated  that  AG  is  equal  to  DF; 
therefore,  since  AB  is  equal  to  DE,  BA  and  AG  are  equal  to  ED 
and  DF;  and  the  angle  BAG  is  equal  to  the  angle  EDF;  wherefore 

D 


the  base  BC  is  equal  (4. 1.)  to  the  base  EF,  and  the  remaining  an- 
gles to  the  remaining  angles:  the  angle  ABG  is  therefore  equal  to 
the  angle  DEF:  and  the  right  angle  ABK  is  equal  to  the  right 


196  THE  ELEMENTS  OF  EUCLID.  BOOK  %1, 

angle  DEN,  whence  the  remaining  angle  CBK  is  equal  to  the  re- 
maining angle  FEN:  for  the  same  reason,  the  angle  BCK  is  equal 
to  the  angle  EFN:  therefore  in  the  two  triangles  BCK,  EFN  there 
are  two  an^s^les  in  one  equal  to  two  angles  in  the  other,  each  to 
each,  and  one  side  equal  to  one  side  adjacent  to  the  equal  angles 
in  each,  viz.  BC  equal  to  EF;  the  other  sides,  therefore,  are  equal 
to  the  other  sides;  BK  then  is  equal  to  EN;  and  AB  is  equal  to 
DE;  wherefore  AB,  BK  are  equal  to  DE,  EN;  and  they  contain 
right  angles:  wherefore  the  base  AK  is  equal  to  the  base  DN: 
and  since  AH  is  equal  to  DM,  the  square  of  AH  is  equal  to  the 
square  of  DM:  but  the  squares  of  AK,  KH  are  equal  to  the  square 
(47.  1.)  of  AH,  because  AKH  is  a  right  angle:  and  the  squares  of 
DN,  NM  are  equal  to  the  square  of  DM,  for  DNM  is  a  right 
angle:  wherefore  the  squares  of  AK,  KH  are  equal  to  the  squares 
of  DN,  NM;  and  of  those  the  square  of  AK  is  equal  to  the  square  of 
DN;  therefore  the  remaining  square  of  KH  is  equal  to  the  remain- 
ing square  of  NM;  and  the  straight  line  KH  to  the  straight  line 
NM:  and  because  HA,  AK  are  equal  to  MD,  DN,  each  to  each, 
and  the  base  HK  to  the  base  MN,  as  has  been  proved;  therefore 
the  angle  HAK  is  equal  (8.  1.)  to  the  angle  MDN.     Q.  E.  D. 

Cor.  From  this  it  is  manifest,  that  if,  from  the  vertices  of  two 
equal  plane  angles,  there  be  elevated  two  equal  straight  lines  con- 
taining equal  angles  with  the  sides  of  the  angles,  each  to  each:  the 
perpendiculars  drawn  from  the  extremities  of  the  equal  straight 
lines  to  the  planes  of  the  first  angles  are  equal  to  one  another. 

Another  Demonstration  of  the  Corollary. 

Let  the  plane  angles  BAC,  EDF  be  equal  to  one  another,  and 
let  AH,  DM,  be  two  equal  straight  lines  above  the  planes  of  the 
angles,  containing  equal  angles  with  BA,  AC;  ED,  DF,  each  to 
each,  viz.  the  angle  HAB,  equal  to  MDE,  and  HAC  equal  to  the 
angle  MDF;  and  from  H,  M  let  HK,  MN  be  perpendiculars  to  the 
planes  BAC,  EDF;  HK  is  equal  to  MN. 

Because  the  solid  angle  at  A  is  contained  by  the  three  plane  an- 
gles BAC,  BAH,  HAC,  which  are,  each  to  each,  equal  to  the  three 
plane  angles  EDF,  EDM,  MDF  containing  the  solid  angle  at  D; 
the  solid  angles  at  A  and  D  are  equal,  and  therefore  coincide  with 
one  another;  to  wit,  if  the  plane  angle  BAC  be  applied  to  the 
plane  angle  EDF,  the  straight  line  AH  coincides  with  DM,  as  was 
shown  in  Prop.  B.  of  this  Book:  and  because  AH  is  equal  to  DM, 
the  point  H  coincides  with  the  point  M;  wherefore  HK,  which  is 
perpendicular  to  the  plane  BAC,  coincides  with  MN  (13.  11.), 
which  is  perpendicular  to  the  plane  EDF,  because  these  planes 
coincide  with  one  another.  Therefore  HK  is  equal  to  MN.  Q. 
E.D. 

PROP.  XXXVI.  THEOR. 

If  three  straight  Hnes  be  proportionals,  the  sohd  pa- 
rallelepiped described  from  all  three  as  its  sides,  is  equal 
to  the  equilateral  parallelepiped   described   from   the 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


197 


mean  proportional,  one  of  the  solid  angles  of  which  is 
contained  by  three  plane  angles  equal,  each  to  each,  to 
the  three  plane  angles  containing  one  of  the  solid  angles 
of  the  other  figure.^ 

Let  A,  B,  C,  be  three  proportionals,  viz.  A  to  B,  as  B  to  C. 
The  solid  described  from  A,  B,  C  is  equal  to  the  equilateral  solid 
described  from  B,  equiangular  to  the  other. 

Take  a  solid  angle  D  contained  by  three  plane  angles  EDF, 
FDG,  GDEj  and  make  each  of  the  straight  lines  ED,  DF,  DG 
equal  to  B,  and  complete  the  solid  parallelepiped  DH.  Make  LK 
equal  to  A,  and  at  the  point  K  in  the  straight  line  LK  make  (26. 
n.)  a  solid  angle  contained  by  the  three  plane  angles  LKM,  MKN, 
NKL,  equal  to  the  angles  EDF,  FDG,  GDE,  each  to  each;  and 
make  KN  equal  to  B,  and  KM  equal  to  C;  and  complete  the  solid 


O 


H 


\ 

h 

M 

\ 

\ 

L 

K 

N 


\ 

N 

F 

\ 

^v 

G 


E 


D 


B 


parallelopiped  KO:  and  because,  as  A  is  to  B,  so  is  B  to  C,  and 
that  A  is  equal  to  LK,  and  B  to  each  of  the  straight  lines  DE, 
DF,  and  C  to  KM;  therefore  LK  is  to  ED,  as  DF  to  KM;  that  is, 
the  sides  about  the  equal  angles  are  reciprocally  proportional; 
therefore  the  parallelogram  LM  is  equal  (14.  6.)  to  EF:  and  be- 
cause EDF,  LKM  are  two  equal  plane  angles,  and  the  two  equal 
straight  lines  DG,  KN  are  drawn  from  their  vertices  above  their 
planes,  and  contain  equal  angles  with  their  sides;  therefore  the 
perpendiculars  from  the  points  G,  N,  to  the  planes  EDF,  LKM 
are  equal  (Cor.  35.  II.)  to  one  another:  therefore  the  solids  KO, 
DH  are  of  the  same  altitude;  and  they  are  upon  equal  bases  LM^ 
EF,  and  therefore  they  are  eqiial  (31.  11.)  to  one  another:  but  the 
solid  KO  is  described  from  the  three  straight  lines  A,  B,  C,  and 
the  solid  DH  from  the  straight  line  B.  If  therefore  three  straight 
lines  &c.    Q.  E.  D. 

PROP.  XXXVn.  THEOR. 

If  four  straight  lines  be  proportionals,  the  similar 
solid  parallelopipeds  similarly  described  from  them  shall 
also  be  proportionals.  And  if  the  similar  parallelopipeds 
similarly  described  from  four  straight  lines  be  propor- 
tionals, the  straight  lines  shall  be  proportionals."^ 


*  See  Note. 


198 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI. 


Let  the  four  straight  lines  AB,  CD,  EF,  GH  be  proportion- 
als, viz.  as  AB  to  CD,  so  EF  to  GH;  and  let  the  similar  paral- 
lelopipeds  AK,  CL,  EM,  GN  be  similarly  described  from  them. 
AK  is  to  CL,  as  EM  to  GN. 

Make  (1 1.  6.)  AB,  CD,  O,  P  continual  proportionals,  as  also  EF, 
GH,  Q,  R;   and  because  as  AB  is  to  CD,  so  EF  to  GH;  and 


■XK 


\i_ 
A. 


\ 


—A 

B 


V— t 

'\    \ 

o 


m: 


that  CD  is  (11.  5.)  to  O,  as  GH  to  Q,  and  O  to  P,  as  Q  to  R; 
therefore,  ex  sequali  (22.  5.),  AB  is  to  P,  as  EF  to  R:  but  as  AB 
to  P,  so  (Cor.  33.  11.)  is  the  solid  AK  to  the  solid  CL;  and  as 
EF  to  R,  so  (Cor.  33.  1 1.)  is  the  solid  EM  to  the  solid  GN;  there- 
fore (11.  5.)  as  the  solid  AK  to  the  solid  CL,  so  is  the  solid  EM 
to  the  solid  GN. 

But  let  the  solid  AK  be  to  the  solid  CL,  as  the  solid  EM  to 
the  solid  GN:  the  straight  line  A B  is  to  ED,  as  EF  to  GH. 

Take  AB  to  CD,  as  EF  to  ST,  and  from  ST  describe  (27.  11.) 
a  solid  parallelopiped  SV  similar  and  similarly  situated  to  either 
of  the  solids  EM,  GN:  and  because  AB  is  to  CD,  as  EF  to 


r\ 


ZZK 


^ 


o 


A, 


J3      C 


I> 


M 


— 


]^f 


Ni 


E 


G        H 


Q 


R 


ST,  and  that  from  AB,  CD  the  solid  parallelopipeds  AK,  CL 
are  similarly  described,  and  in  like  manner  the  solids  EM,  SV 
from  the   straight   lines  EF,   ST;   therefore  AK   is  to  CL,  as 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


199 


EM  to  SV:  but,  by  the  hypothesis,  AK  is  to  CL,  as  EM  to  GN; 
therefore  GN  is  equal  (9.  5.)  to  SV:  but  it  is  likewise  similar  and 
similarly  situated  to  SV;  therefore  the  planes  which  contain  the 
solids  GN,  SV  are  similar  and  equal,  and  their  homologous  sides 
GH,  ST  equal  to  one  another:  and  because  as  AB  to  CD  so  EF 
to  ST,  and  that  ST  is  equal  to  GH,  AB  is  to  CD,  as  EF  to  GH. 
Therefore,  if  four  straight  lines,  Sec.     Q.  E.  D. 


PROP.  XXXVni.  THEOR. 


u 


'If  a  plane  be  perpendicular  to  another  plane,  and  a 
straight  line  be  drawn  from  a  point  in  one  of  the  planes 
perpendicular  to  the  other  plane,  this  straight  lae  shall 
fall  on  the  common  section  of  the  planes."* 

"Let  the  plane  CD  be  perpendicular  to  the  plane  AB,  and  let 
AD  be  their  common  section^  if  any  point  E  be  taken  in  the 
plane  CD,  the  perpendicular  drawn  from  E  to  the  plane  AB  shall 
fall  on  AD. 

"  For,  if  it  does  not,  let  it,  if  possible,  fall  elsewhere,  as  EF; 
and  let  it  meet  the  plane  AB  in  the  point  F;  and  from  F  draw 
(12.  1.)  in  the  plane  AB  a  perpendicular  FG  to  DA,  which  is  also 
perpendicular  (4.  def.  11.)  to  the  plane  CD;  and  join  EG:  then 
because  FG  is  perpendicular  to  the  C 
plane  CD,  and  the  straight  line  EG, 
which  is  in  that  plane,  meets  it; 
therefore  FGE  is  a  right  angle  (3. 
def.  11.):  but  EF  is  also  at  right  an- 
gles to  the  plane  AB;  and  therefore 
EFG  is  a  right  angle:  wherefore 
two  of  the  angles  of  the  triangle 
EFG  are  equal  together  to  two  right 
angles;  which  is  absurd:  therefore  the  perpendicular  from  the 
point  E  to  the  plane  AB,  does  not  fall  elsewhere  than  upon  the 
straight  line  AD;  it  therefore  falls  upon  it.  If  therefore  a  plane," 
&c.     Q.  E.  D. 


A 


B 


PROP.  XXXIX.   THEOR. 


In  a  solid  parallelepiped,  if  the  sides  of  two  of  the 
opposite  planes  be  divided  each  into  two  equal  parts, 
the  common  section  of  the  planes  passing  through  the 
points  of  division,  and  the  diameter  of  the  solid  paral- 
lelopiped  cut  each  other  into  two  equal  parts.* 


See  Note, 


200 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XI. 


K 


E 


Let  the  sides  of  the  D 
opposite  planes  CF, 
AH  of  the  solid  paral- 
lelopiped  AF,  be  di- 
vided each  into  two 
equal  parts  in  the 
points  K,  L,  M,N5  X, 
O,  P,  R;  and  join  KL, 
MN,  XO,  PR:  and 
because  DK,  CL  are 
equal  and  parallel,  KL 
is  parallel  (33.  1.)  to 
DC:  for  the  same  rea-  B 
son,  MN  is  parallel  to 
BA:  and  BA  is  paral- 
lel to  DC;  therefore 
because  KL,  BA,  are  A  N  G 

each  of  them  parallel  to  DC,  and  not  in  the  same  plane  with  it, 
KL  is  parallel  (9.  11.)  to  BA:  and  because  KL,  MN  are  each  of 
them  parallel  to  BA,  and  not  in  the  same  plane  with  it,  KL  is  pa- 
rallel (9.  11.)  to  MN;  wherefore  KL,  MN  are  in  one  plane.  In 
like  manner,  it  may  be  proved,  that  XO,  PR  are  in  one  plane. 
Let  YS  be  the  common  section  of  the  planes  KN,  XR;  and  DG 
the  diameter  of  the  solid  parallelopiped  AF:  YS  and  DG  do  meet, 
and  cut  one  another  into  two  equal  parts. 

Join  DY,  YE,  BS,  SG.  Because  DX  is  parallel  to  OE,  the  al- 
ternate angles  DXY,  YOE  are  equal  (29.  1.)  to  one  another:  and 
because  DX  is  equal  to     D  K  F 

OE,andXYtoYO,  and 
contain  equal  angles, 
the  base  DY  is  equal 

(4.  1.)  to  the  base  YE,  XX  ~  I         I      V^l      ^'T^"'--^    E 

and  the  other  angles 
are  equal;  therefore 
the  angle  XYD  is  equal 
to  the  angle  OYE,  and 
DYE  is  a  straight  (14. 
1.)  line:  for  the  same 
reason  BSG  is  a  straight 
line,  and  BS  equal  to 
SG:  and  because  CA 
is  equal  and  parallel  to 
DB,  and  also  equal  and 
parallel  to  EG,  there- 
fore  DB   is   equal  and 

parallel  (9.  1 1.)  to  EG:  and  DE,  BG  join  their  extremities;  there- 
fore DE  is  equal  and  parallel  (33.  1.)  to  BG:  and  DG,  YS  are 
drawn  from  points  in  the  one,  to  points  in  the  other;  and  are 
therefore  in  one  plane:  whence  it  is  manifest,  that  DG,  YS  must 
meet  one  another;  let  them  meet  in  T:  and  because  DE  is  paral- 
lel to  BG,  the  alternate  angles  EDT,  BGT  are  equal  (29.  1.);  and 
the  angle  DTY  is  equal  (15.  1.)  to  the  angle  GTS:  therefore  in 


BOOK  XI. 


THE  ELEMENTS  OF  EUCLID. 


201 


the  triangles  DTY,  GTS  there  are  two  angles  in  the  one  equal  to 
two  angles  in  the  other,  and  one  side  equal  to  one  side,  opposite 
to  two  of  the  equal  angles,  viz.  DY  to  GS;  for  they  are  the  halves 
of  DE,  BG:  therefore  the  remaining  sides  are  equal  (26.  1.)  each 
to  each.  Wherefore,  DT  is  equal  to  TG,  and  YT  equal  to  TS. 
Wherefore,  if  in  a  solid,  &c.     Q.  E.  D. 

PROP.  XL.  THEOR. 

If  there  be  two  triangular  prisms  of  the  same  alti- 
tude, the  base  of  one  of  which  is  a  parallelogram,  and 
the  base  of  the  other  a  triangle;  if  the  parallelogram 
be  double  of  the  triangle,  the  prisms  shall  be  equal  to 
one  another. 

Let  the  prisms  ABCDEF,  GHKLMN  be  of  the  same  altitude, 
the  first  whereof  is  contained  by  the  two  triangles  ABE,  CDF, 
and  the  three  parallelograms  AD,  DE,  EC;  and  the  other  by  the 
two  triangles  GHK,  LMN,  and  the  three  parallelograms  LH,  HN, 
NG;  and  let  one  of  them  have  a  parallelogram  AF,  and  the  other 
a  triangle  GHK  for  its  base;  if  the  parallelogram  AF  be  double 
of  the  triangle  GHK,  the  prism  ABCDEF  is  equal  to  the  prism 
GHKLMN. 

Complete  the  solids  AX,  GO;  and  because  the  parallelogram 
AF  is  double  of  the  triangle  GHK;  and  the  parallelogram  HK 


B 

D 

iV- 

K 

\ 

A 

N^ 

A 

X 


H 


K 


double  (34.  1.)  of  the  same  triangle;  therefore  the  parallelogram 
AF  is  equal  to  HK.  But  solid  parallelopipeds  upon  equal  bases, 
and  of  the  same  altitude,  are  equal  (31.  11.)  to  one  another. 
Therefore  the  solid  AX  is  equal  to  the  solid  GO;  and  the  prism 
ABCDEF  is  half  (28.  11.)  of  the  solid  AX;  and  the  prism 
GHKLMN  half  (28.  11.)  of  the  solid  GO.  Therefore  the  prism 
ABCDEF  is  equal  to  the  prism  GHKLMN.  Wherefore,  if  there 
be  two,  8cc.     Q.  E.  D. 


2C 


THE 


EI.EMENTS  OF  EUCLID, 


BOOK  XII. 


LEMMA  I. 


K- 


H 


Which  is  the  first  proposition  of  the  tenth  book,  and  is  necessary 
to  some  of  the  propositions  of  this  book. 

If  from  the  greater  of  two  unequal  magnitudes,  there 
be  taken  more  than  its  half,  and  from  the  remainder 
more  than  its  half,  and  so  on:  there  shall  at  length  re- 
main a  magnitude  less  than  the  least  of  the  proposed 
magnitudes."^ 

Let  AB  and  C  be  two  unequal  magnitudes,  of 
which  AB  is  the  greater.     If  from  AB  there  be  D 

taken  more  than  its  half,  and  from  the  remainder 
more  than  its  half,  and  so  on^  there  shall  at  length 
remain  a  magnitude  less  than  C. 

For  C  may  be  multiplied,  so  as  at  length  to  be- 
come greater  than  AB.  Let  it  be  so  multiplied, 
and  let  DE  its  multiple  be  greater  than  AB,  and 
let  DE  be  divided  into  DF,  FG,  GE,  each  equal 
to  C.  From  AB  take  BH  greater  than  its  half, 
and  from  the  remainder  AH   take  HK  greater  G 

than  its  half,  and  so  on,  until  there  be  as  many- 
divisions  in  AB  as  there  are  in  DE:  and  let  the 
divisions  in  AB  be  AK,  KH,  HB;  and  the  divi- 
sions in  ED  be  DF,  FG,  GE.  And  because  DE 
is  greater  than  AB,  and  that  EG  taken  from  DE  is  not  greater 
than  its  half,  but  BH  taken  from  AB  is  greater  than  its  halfj 
therefore  the  remainder  GD  is  greater  than  the  remainder  HA. 
Again,  because  GD  is  greater  than  HA,  and  that  GF  is  not 
greater  than  the  half  of  GD,  but  HK  is  greater  than  the  half  of 
HA;  therefore  the  remainder  FD  is  greater  than  the  remainder 


B  C 


"  See  Note. 


BOOK  XII. 


THE  ELEMENTS  OF  EUCLID. 


203 


AK.    And  FD  is  equal  to  C,  therefore  C  is  greater  than  AKj  that 
is,  AK,  is  less  than  C.     Q.  E.  D. 

And  if  only  the  halves  be  taken  away,  the  same  thing  may  in 
the  same  way  be  demonstrated. 


PROP.  I.  THEOR. 

Similar  polygons  inscribed   in  circles   are  to  one 
another  as  the  squares  of  their  diameters. 

Let  ABCDE,  FGHKL  be  two  circles,  and  in  them  the  similar 
polygons  ABCDE,  FGHKLj  and  let  BM,  GN  be  the  diameters 
of  the  circles;  as  the  square  of  BM  is  to  the  square  of  GN,  so  is 
the  polygon  ABCDE  to  the  polygon  FGHKL. 
^  Join  BE,  AM,  GL,  FN:  and  because  the  polygon  ABCDE  is 
similar  to  the  polygon  FGHKL,  and  similar  polygons  are  divided 
into  similar  triangles;  the  triangles  ABE,  FGL  are  similar  and 

A  F 


equiangular  (6.  6.);  and  therefore  the  angle  AEB  is  equal  to  the 
angle  FLG:  but  AEB  is  equal  (2 1 .  3.)  to  AMB,  because  they  stand 
upon  the  same  circumference;  and  the  angle  FLG  is,  for  the  same 
reason,  equal  to  the  angle  FNG:  therefore  also  the  angle  AMB  is 
equal  to  FNG:  and  the  right  angle  BAM  is  equal  to  the  right 
(3L  3.)  angle  GFN;  wherefore  the  remaining  angles  in  the  trian- 
gles ABM,  FGN  are  equal,  and  they  are  equiangular  to  one  ano- 
ther: therefore  as  BM  to  GN,  so  (4.  6.)  is  BA  to  GF;  and  therefore 
the  duplicate  ratio  of  BM  to  GN,  is  the  same  (10.  dcf.  5.  and  22.  5.) 
with  the  duplicate  ratio  of  BA  to  GF:  but  the  ratio  of  the  square 

F 


B 


of  BM  to  the  square  of  GN  is  the  duplicate  (20.  6.)  ratio  of  that 
which  BM  has  to  GN;  and  the  ratio  of  the  polygon  ABCDE  to 
the  polygon  FGHKL  is  the  duplicate  (20.  6.)  of  that  which  BA 
has  to  GF;  therefore,  as  the  square  of  BM   to  the  square  of  GN, 


204  THE    ELEMENTS    OF    EUCLID.  BOOK  XII. 

SO  is  the  polygon  ABCDE,  to  the  polygon  FGHKL.     Wherefore, 
similar  polygons,  &c.  Q.  E.  D. 

PROP.  II.  THEOR. 

Circles  are  to  one  another  as  the  squares  of  their 
diameters.* 

Let  ABCD,  EFGH  be  two  circles,  and  BD,  FH  their  diame- 
ters: as  the  square  of  BD  to  the  square  of  FH,  so  is  the  circle 
ABCD,  to  the  circle  EFGH. 

For,  if  it  be  not  so,  the  square  of  BD  shall  be  to  the  square  of 
FH,  as  the  circle  ABCD  is  to  some  space  either  less  than  the 
circle  EFGH,  or  greater  than  it.f  First  let  it  be  to  a  space  S  less 
than  the  circle  EFGH;  and  in  the  circle  EFGH  describe  the  square 
EFGH:  this  square  is  greater  than  half  of  the  circle  EFGH;  be- 
cause if,  through  the  points  E,  F,  G,  H,  there  be  drawn  tangents  to 
the  circle,  the  square  EFGH  is  half  (41.  1.)  of  the  square  descri- 
bed about  the  circle;  and  the  circle  is  less  than  the  square  described 
about  it;  therefore  the  square  EFGH  is  greater  than  half  of  the 
circle.  Divide  the  circumferences  EF,  FG,  GH,  HE,  each  into 
two  equal  parts  in  the  points  K,  L,  M,  N,  and  join  EK,  KF,  FL, 
LG,  GM,  MH,  HN,  NE:  therefore  each  of  the  triangles  EKF, 
FLG,  GMH,  HNE  is  greater  than  half  of  the  segment  of  the  cir- 
cle it  stands  in;  because,  if  straight  lines  touching  the  circle  be 
drawn  through  the  points  K,  L,  M,  N,  and  parallelograms  upon 
the  straight  lines  EF,  FG,  GH,  HE  be  completed;  each  of  the  tri- 
angles EKF,  FLG,  GMH,  HNE  shaU  be  the  half  (41.  1.)  of  the 
parallelogram  in  which  it  is:  but  every  segment  is  less  than  the 
parallelogram  in  which  it  is:  wherefore  each  of  the  triangles  EKF, 
FLG,  GMH,  HNE  is  greater  than  half  the  segment  of  the  circle 
which  contains  it:  and  if  these  circumferences  before  named  be 
divided  each  into'two  equal  parts,  and  their  extremities  be  joined 
A 


by  straight  lines,  by  continuing  to  do  this,  there  will  at  length  re- 
main segments  of  the  circle,  which,  together,  shall  be  less  than  the 

*  See  Note. 

t  For  there  is  some  square  equal  to  the  circle  ABCD;  let  P  be  the  side  of  it; 
and  to  three  straight  lines  BD,  FH,  and  P,  there  can  be  a  fourth  proportional ; 
let  this  be  Q:  therefore  the  squares  of  these  four  straight  lines  are  proportionals, 
that  is,  to  the  squares  of  BD,  Ffl,  and  the  circle  ABCD,  it  is  possible  there  may 
be  a  fourth  proportional.  Let  this  be  S.  And  in  liiie  manner,  are  to  be  under- 
stood some  things  in  some  of  the  following  propositions. 


BOOK  XII. 


THE  ELEMENTS  OF  EUCLID. 


205 


excess  of  the  circle  EFGH  above  the  space  S:  because,  by  the 
preceding  lemma,  if  from  the  greater  of  two  unequal  magnitudes 
there  be  taken  more  than  its  half,  and  from  the  remainder  more 
than  its  half,  and  so  on,  there  shall  at  length  remain  a  magnitude 
less  than  the  least  of  the  proposed  magnitudes.  Let  then  the  seg- 
ments EK,  KF,  FL,  LG,  GM,  MH,  HN,  NE,  be  those  that  remain 
and  are  together  less  than  the  excess  of  the  circle  EFGH  above 
S:  therefore  the  rest  of  the  circle,  viz.  the  polygon  EKFLGMHN, 
is  greater  than  the  space  S.  Describe  likewise  in  the  circle  ABCD 
the  polygon  AXBOCPDR  similar  to  the  polygon  EKFLGMHN:  as 
therefore,  the  square  of  BD  is  to  the  square  of  FH,  so  (1.  12.)  is 
the  polygon  AXBOCPDR  to  the  polygon  EKFLGMHN:  but  the 
square  of  BD  is  also  to  the  square  of  FH,  as  the  circle  ABCD  is 
to  the  space  S:  therefore  as  the  circle  ABCD  is  to  the  space  S,  so 
is  (1 1. 5.)  the  polygon  AXBOCPDR  to  the  polygon  EKFLGMHN: 
but  the  circle  ABCD  is  greater  than  the  polygon  contained  in  it: 

A 


B 


D 


wherefore  the  space  S  is  greater  (14.  5.)  than  the  polygon 
EKFLGMHN:  but  it  is  likewise  less,  as  has  been  demonstrated: 
which  is  impossible.  Therefore  the  square  of  BD  is  not  to  the 
square  of  FH,  as  the  circle  ABCD  is  to  any  space  less  than  the 
circle  EFGH.  In  the  same  manner,  it  may  be  demonstrated,  that 
neither  is  the  square  of  FFI  to  the  square  of  BD,  as  the  circle 
EFGH  is  to  any  space  less  than  the  circle  ABCD.  Nor  is  the 
square  of  BD  to  the  square  of  FH,  as  the  circle  ABCD  is  to  any 
space  greater  than  the  circle  EFGH:  for,  if  possible,  let  it  be  so 
to  T,  a  space  greater  than  the  circle  EFGH:  therefore,  inversely, 
as  the  square  of  FH  to  the  square  of  BD,  so  is  the  space  T  to  the 
circle  ABCD.     But  as  the  space  T*  is  to  the  circle  ABCD,  so  is 

*  For,  as  in  the  foregoing  note,  at  t  it  was  explained  how  it  was  possible  there 
could  be  a  fourth  proportional  to  the  squares  of  BD,  FH,  and  the  circle  ABCD, 
which  was  named  S.  So  in  like  manner  there  can  be  a  fourth  proportional  to 
this  other  space  named  T,  and  the  circles  ABCD,  EFGH.  And  the  like  is  to  be 
understood  in  some  of  the  following  propositions. 


■^ 


206 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XII. 


the  circle  EFGH  to  some  space,  which  must  be  less  (14.  5.)  than 
the  circle  ABCD,  because  the  space  T  is  greater  by  hypothesis, 
than  the  circle  EFGH.  Therefore  as  the  square  of  FH  is  to  the 
square  of  BD,  so  is  the  circle  EFGH  to  a  space  less  than  the 
circle  ABCD,  which  has  been  demonstrated  to  be  impossible: 
therefore  the  square  of  BD  is  not  to  the  square  of  FH,  as  the  cir- 
cle ABCD  is  to  any  space  greater  than  the  circle  EFGH:  and  it 
has  been  demonstrated,  that  neither  is  the  square  of  BD  to  the 
square  of  FH,  as  the  circle  ABCD  is  to  any  space  less  than  the  cir- 
cle EFGH:  wherefore,  as  the  square  of  BD  to  the  square  of  FH, 
so  is  the  circle  ABCD  to  the  circle  EFGH.*  Circles  therefore 
are,  &c.     Q.  E.  D. 

PROP.  HI.  THEOR. 

Every  pyramid  having  a  triangular  base,  may  be  di- 
vided into  two  equal  and  similar  pyramids  having  tri- 
angular bases,  and  which  are  similar  to  the  whole  pyra- 
mid; and  into  two  equal  prisms  which  together  are 
greater  than  half  of  the  whole  pyramid.t 

Let  there  be  a  pyramid  of  which  the  base  is  the  triangle  ABC, 
and  its  vertex  the  point  D:  the  pyramid  ABCD  may  be  divided 
into  two  equal  and  similar  pyramids  having 
triangular  bases,  and  similar  to  the  whole; 
and  into  two  equal  prisms  which  together  are 
greater  than  half  of  the  whole  pyramid. 

Divide  AB,  BC,  CA,  AD,  DB,  DC,  each 
into  two  equal  parts  in  the  points  E,  F,  G, 
H,  K,  L,  and  join  EH,  EG,  GH,  HK,  KL, 
LH,  EK,  KF,  FG.  Because  AE  is  equal  to 
EB,  and  AH  to  HD,  HE  is  parallel  (2.  6.) 
to  DB;  for  the  same  reason,  HK  is  parallel 
to  AB:  therefore  HEBK  is  a  parallelogram, 
and  HK  equal  (34.  1.)  to  EB:  but  EB  is 
equal  to  AE;  therefore  also  AE  is  equal  to 
HK:  and  AH  is  equal  to  HD;  wherefore 
EA,  AH  are  equal  to  KH,  HD,  each  to  each; 
and  the  angle  EAH  is  equal  (29.  1.)  to  the     I^  F  C 

angle  KHD;  therefore  the  base  EH  is  equal  to  the  base  KD,  and 
the  triangle  AEH  equal  (4.  I.)  and  similar  to  the  triangle  HKD: 
for  the  same  reason,  the  triangle  AGH  is  equal  and  similar  to 
the  triangle  HLD:  and  because  the  two  straight  lines  EH,  HG, 
which  meet  one  another,  are  parallel  to  KD,  DL  that  meet 
one  another,  and  are  not  in  the  same  plane  with  them,  they  con- 
tain equal  (10.  11.)  angles;  therefore  the  angle  EHG  is  equal  to 
the  angle  KDL.  Again,  because  EH,  HG,  are  equal  to  KD, 
DL,  each  to  each,  and  the  angle  EHG  equal  to  the  angle  KDL; 
therefore  the  base  EG  is  equal  to  the  base  KL;  and  the  triangle 

*  Because  as  a  fourth  proportional  to  the  squares  of  BD,  FH,  and  the  circle 
ABCD  is  possible,  and  that  it  can  neither  be  less  nor  greater  than  the  circle 
EFGH,  it  must  be  equal  to  it. 

t  See  Note. 


BOOK  XII. 


THE    ELEMENTS    OF   EUCLID. 


2or 


EHG  equal  (4.  1.)  and  similar  to  the  triangle  KDL;  for  the 
same  reason  the  triangle  AEG  is  also  equal  and  similar  to  the 
triangle  HKL.  Therefore  the  pyramid  of  which  the  base  is  the 
triangle  AEG,  and  of  which  the  vertex  is  the  point  H,  is  equal 
(C.  11.)  and  similar  to  the  pyramid  the  base  of  which  is  the 
triangle  KHL,  and  vertex  the  point  D:  and  D 

because  HK  is  parallel  to  AB  a  side  of  the  tri- 
angle ADB,  the  triangle  ADB,  is  equiangu- 
lar to  the  triangle  HDK,  and  their  sides  are 
proportionals  (4.  6.):  therefore  the  triangle 
ADB  is  similar  to  the  triangle  HDK:  and 
for  the  same  reason,  the  triangle  DBC  is  si- 
milar to  the  triangle  DKLj  and  the  triangle 
ADC  to  the  triangle  HDLj  and  also  the  tri- 
angle ABC  to  the  triangle  AEG:  but  the  tri- 
angle AEG  is  similar  to  the  triangle  HKL,  as 
before  was  proved;  therefore  the  triangle 
ABC  is  similar  (21.  6.)  to  the  triangle  HKL. 
And  the  pyramid  of  which  the  base  is  the 
triangle  ABC,  and  vertex  the  point  D,  is 
therefore  similar  (B.  11.  and  11.  def.  11.)  to 
the  pyramid  of  which  the  base  is  the  triangle 

HKL,  and  vertex  the  same  point  D:  but  the  pyramid  of  which  the 
base  is  the  triangle  HKL,  and  vertex  the  point  D,  is  similar,  as  has 
been  proved,  to  the  pyramid  the  base  of  which  is  the  triangle 
AEG,  and  vertex  the  point  H:  wherefore  the  pyramid,  the  base  of 
which  is  the  triangle  ABC,  and  vertex  the  point  D,  is  similar  to 
the  pyramid  of  which  the  base  is  the  triangle  AEG  and  vertex  H: 
therefore  each  of  the  pyramids  AEGH,  HKLD  is  similar  to  the 
whole  pyramid  ABCD:  and  because  BF  is  equal  to  FC,  the  paral- 
lelogram EBFG  is  double  (41.  1.)  of  the  triangle  GFC:  but  when 
there  are  two  prisms  of  the  same  altitude,  of  which  one  has  a  pa- 
rallelogram for  its  base,  and  the  other  a  triangle  that  is  half  of 
the  parallelogram,  these  prisms  are  equal  (40.  11.)  to  one  another; 
therefore  the  prism  having  the  parallelogram  EBFG  for  its  base, 
and  the  straight  line  KH  opposite  to  it,  is  equal  to  the  prism  hav- 
ing the  triangle  GFC  for  its  base,  and  the  triangle  HKL  opposite 
to  it;  for  they  are  of  the  same  altitude,  because  they  are  between 
the  parallel  (15.  11.)  planes  ABC,  HKL:  and  it  is  manifest  that 
each  of  these  prisms  is  greater  than  either  of  the  pyramids  of 
which  the  triangles  AEG,  HKL  are  the  bases,  and  the  vertices 
the  points  H,  D;  because  if  E,  F  be  joined,  the  prism  having 
the  parallelogram  EBFG  for  its  base,  and  KH  the  straight  line 
opposite  to  it,  is  greater  than  the  pyramid  of  which  the  base 
is  the  triangle  EBF,  and  vertex  the  point  K;  but  this  pyramid  is 
equal  (C.  11.)  lo  the  pyramid  the  base  of  which  is  the  triangle 
AEG,  and  vertex  the  point  H;  because  they  are  contained  by  equal 
and  similar  planes:  wherefore  the  prism  having  the  parallelogram 
EBFG  for  its  base,  and  opposite  side  KH,  is  greater  than  the 
pyramid  of  which  the  base  is  the  triangle  AEG,  and  vertex  the 
point  H:  and  the  prism  of  which  the  base  is  the  parallelogram 
EBFG,  and  opposite  side  KH,  is  equal  to  the  prism  having  the 


208  THE  ELEMENTS  OF  EUCLID.  BOOK  XII. 

triangle  GFC  for  its  base,  and  HKL  the  triangle  opposite  to  it; 
and  the  pyramid  of  which  the  base  is  the  triangle  AEG,  and  ver- 
tex H,  is  equal  to  the  pyramid  of  which  the  base  is  the  triangle 
HKL,  and  vertex  D:  therefore  the  two  prisms  before  mentioned 
are  greater  than  the  two  pyramids  of  which  the  bases  are  the  tri- 
angles AEG,  HKL,  and  vertices  the  points  H,  D.  Therefore  the 
whole  pyramid  of  which  the  base  is  the  triangle  ABC,  and  vertex 
the  point  D,  is  divided  into  two  equal  pyramids  similar  to  one 
another,  and  to  the  whole  pyramid;  and  into  two  equal  prisms; 
and  the  two  prisms  are  together  greater  than  half  of  the  whole 
pyramid.     Q.  E.  D. 

PROP.  IV.  THEOR. 

If  there  be  two  pyramids  of  the  same  altitude,  upon 
triangular  bases,  and  each  of  them  be  divided  into  two 
equal  pyramids  similar  to  the  whole  pyramid,  and  also 
into  two  equal  prisms;  and  if  each  of  these  pyramids  be 
divided  in  the  same  manner  as  the  first  two,  and  so  on: 
as  the  base  of  one  of  the  first  two  pyramids  is  to  the 
base  of  the  other,  so  shall  all  the  prisms  in  one  of  them 
be  to  all  the  prisms  in  the  other,  that  are  produced  by 
the  same  number  of  divisions.* 

Let  there  be  two  pyramids  of  the  same  altitude  upon  the  trian- 
gular bases  ABC,  DEF,  and  having  their  vertices  in  the  points  G, 
H;  and  let  each  of  them  be  divided  into  two  equal  pyramids  si- 
milar to  the  whole,  and  into  two  equal  prisms;  and  let  each  of  the 
pyramids  thus  made  be  conceived  to  be  divided  in  the  like  man- 
ner, and  so  on:  as  the  base  ABC  is  to  the  base  DEF,  so  are  all  the 
prisms  in  the  pyramid  ABCG,  to  all  the  prisms  in  the  pyramid 
DEFH  made  by  the  same  number  of  divisions. 

Make  the  same  construction  as  in  the  foregoing  proposition: 
and  because  BX  is  equal  to  XC,  and  AL  to  LC;  therefore  XL  is 
parallel  (2.  6.)  to  AB,  and  the  triangle  ABC  similar  to  the  triangle 
LXC:  for  the  same  reason,  the  triangle  DEF  is  similar  to  RVF: 
and  because  BC  is  double  of  CX,  and  EF  double  of  FV,  therefore 
BC  is  to  CX,  as  EF  to  FV:  and  upon  BC,  CX  are  described  the 
similar  and  similarly  situated  rectilineal  figures  ABC,  LXC;  and 
upon  EF,  FV,  in  like  manner,  are  described  the  similar  figures 
DEF,  RVF:  therefore,  as  the  triangle  ABC  is  to  the  triangle 
LXC,  so  (22.  6.)  is  the  triangle  DEF  to  the  triangle  RVF,  and, 
by  permutation,  as  the  triangle  ABC  to  the  triangle  DEF,  so  is 
the  triangle  LXC  to  the  triangle  RVF:  and  because  the  planes 
ABC,  OMN,  as  also  the  planes  DEF,  STY  are  parallel  (5.  11.), 
the  perpendiculars  drawn  from  the  points  G,  H  to  the  bases  ABC, 
DEF,  which,  by  the  hypothesis,  are  equal  to  one  another,  shall  be 
cut  each  into  two  equal  (17.  11.)  parts  by  the  planes  OMN,  STY, 
because  the  straight  lines  GC,  HF  are  cut  into  two  equal  parts 

^  See  Note. 


BOOK  XII. 


THE  ELEMENTS  OF  EUCLID. 


209 


in  the  points  N,  Y  by  the  same  planes:  therefore  the  prisms 
LXCOMN,  RVFSTY'are  of  the  same  altitude;  and  therefore  as 
the  base  LXC  to  the  base  RVF:  that  is,  as  the  triangle  ABC  to 
the  triangle  DEF,  so  (Cor.  32.  11.)  is  the  prism  having  the  tri- 
angle LXC  for  its  base,  and  OMN  the  triangle  opposite  to  it,  to 
the  prism  of  which  the  base  is  the  triangle  RVF,  and  the  oppo- 
site triangle  STY:  and  because  the  two  prisms  in  the  pyramid 
ABCG  are  equal  to  one  another,  and  also  the  two  prisms  in  the 
pyramid  DEFH  equal  to  one  another,  as  the  prism  of  which  the 
base  is  the  parallelogram  KBXL  and  opposite  side  MO,  to  the 
prism  having  the  triangle  LXC  for  its  base,  and  OMN  the  triangle 
opposite  to  it,  so  is  the  prism  of  which  the  liase  (7.  5.)  is  the  pa- 
rallelogram PEVR,  and  opposite  side  TS,  to  the  prism  of  which 
the  base  is  the  triangle  RVF,  and  opposite  triangle  STY.  There- 
fore componcndo,  as  the  prisms  KBXLMO,  LXCOMN  together 
are    unto    the    prism    LXOMN,  so  are    the   prisms    PEVRTS, 

G  H 


B  X  C         E  V  F 

RVFSTY,  to  the  prism  RVFSTY-  and,  permutando,  as  the 
prisms  KBXLMO,  LXCOMN  are  to  the  prisms  PEVRTS, 
RVFSTY,  so  is  the  prism  LXCOMN  to  the  prism  RVFSTY:  but 
as  the  prism  LXCOMN  to  the  prism  RVFSTY,  so  is,  as  has  been 
proved,  the  base  ABC  to  the  base  DEF:  therefore,  as  the  base 
ABC  to  the  base  DEF,  so  are  the  two  prisms  in  the  pyramid 
ABCG  to  the  two  prisms  in  the  pyramid  DEFH:  and  likewise  if 
the  pyramids  now  made,  for  example,  the  two  OMNG,  STYH,  be 
divided  in  the  same  manner;  as  the  base  OMN  is  to  the  base  STY, 
so  shall  the  two  prisms  in  the  pyramid  OMNG  be  to  the  two 
prisms  in  the  pyramid  STYH:  but  the  base  OMN  is  to  the  base 
STY,  as  the  base  ABC  to  the  base  DEF;  therefore,  as  the  base 
ABC  to  the  base  DEF,  so  are  the  two  prisms  in  the  pyramid 
ABCG  to  the  two  prisms  in  the  pyramid  DEFH;  and  so  are  the 
two  prisms  in  the  pyramid  OMNG  to  the  two  pi"isms  in  the  py- 
ramid STYH;  and  so  are  all  four  to  all  four:  and  the  same  thing 
may  be  shown  of  the  prisms  made  by  dividing  the  pyramids  AKLO 
and  DPRS,  and  of  all  made  bv  the  same  number  of  divisions, 
Q.E.D. 

2  D 


210 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XII  . 


PROP.  V.  THEOR. 

Pyramids  of  the  same  altitude,  which  have  triangular 
bases,  are  to  one  another  as  their  bases.* 

Let  the  pyramids  of  which  the  triangles  ABC,  DEF  are  the 
bases,  and  of  which  the  vertices  are  the  points  G,  H,  be  of  the 
same  altitude;  as  the  base  ABC,  to  the  base  DEF,  so  is  the  pyra- 
mid ABCG  to  the  pyramid  DEFH. 

For,  if  it  be  not  so,  the  base  ABC  must  be  to  the  base  DEF,  as 
the  pyramid  ABCG  to  a  solid  either  less  than  the  pyramid  DEFH, 
or  greater  than  it.f  First,  let  it  be  to  a  solid  less  than  it,  viz.  to 
the  solid  Q:  and  divide  the  pyramid  DEFH  into  two  equal  pyra- 
mids, similar  to  the  whole,  and  into  two  equal  prisms:  therefore 
these  two  prisms  are  greater  (3.  12.)  than  the  half  of  the  whole* 
pyramid.  And  again,  let  the  pyramids  made  by  this  division  be 
in  like  manner  divided,  and  so  on,  until  the  pyramids  which  re- 
main undivided  in  the  pyramid  DEFH  be,  all  of  them  together, 
less  than  the  excess  of  the  pyramid  DEFH  above  the  solid  Q:  let 
these,  for  example,  be  the  pyramids  DPRS,  STYH:  therefore 
the  prisms,  which  make  the  rest  of  the  pyramid  DEFH,  are 
greater  than  the  solid  Q:  divide  likewise  the  pyramid  ABCG  in 
the  same  manner,  and  into  as  many  parts,  as  the  pyramid  DEFH: 
therefore,  as  the  base  ABC  to  the  base  DEF,  so  (4.  12.)  are  the 
prisms  in  the  pyramid  ABCG  to  the  prisms  in  the  pyramid  DEFH: 
but  as  the  base  ABC  to  the  base  DEF,  so,  by  hypothesis,  is  the 
pyramid  ABCG  to  the  solid  Q;  and  therefore,  as  the  pyramid 
ABCG  to  the  solid  Q,  so  are  the  prisms  in  the  pyramid  ABCG, 
to  the  prisms  in  the  pyramid  DEFH;  but  the  pyramid  ABCG  is 
greater  than  the  prisms  contained  in  it;  wherefore  (14.  5.)  also 

G 


B 


C     E 


V 


*  See  Note, 
th^  iT"^  "^^^  ^^  explained  the  same  way  as  at  the  note  i  in  proposition  2  in 


BOOK  XII. 


THE  ELEMENTS  OF  EUCLID. 


211 


the  solid  Q  is  greater  than  the  prisms  in  the  pyramid  DEFH. 
But  it  is  also  less,  which  is  impossible.  Therefore  the  base  ABC 
is  not  to  the  base  DEF,  as  the  pyramid  ABCG  to  any  solid  which 
is  less  than  the  pyramid  DEFH.  In  the  same  manner  it  may  be 
demonstrated,  that  the  base  DEF  is  not  to  the  base  ABC,  as  the 
pyramid  DEFH  to  any  solid  which  is  less  than  the  pyramid 
ABCG.  Nor  can  the  base  ABC  be  to  the  base  DEF,  as  the  py- 
ramid ABCG  to  any  solid  which  is  greater  than  the  pyramid 
DEFH.  For,  if  it  be  possible,  let  it  be  so  to  a  greater,  viz.  the 
solid  Z.  And  because  the  base  ABC  is  to  the  base  DEF,  as  the 
pyramid  ABCG  to  the  solid  Z;  by  inversion,  as  the  base  DEF  to 
the  base  ABC,  so  is  the  solid  Z  to  the  pyramid  ABCG.  But  as 
the  solid  Z  is  to  the  pyramid  ABCG,  so  is  the  pyramid  DEFH  to 
some  solid,*  which  must  be  less  (14.  5.)  than  the  pyramid  ABCG, 
because  the  solid  Z  is  greater  than  the  pyramid  DEFH.  And 
therefore,  as  the  base  DEF  to  the  base  ABC,  so  is  the  pyramid 
DEFH  to  a  solid  less  than  the  pyramid  ABCG;  the  contrary  to 
which  has  been  proved.  Therefore  the  base  ABC  is  not  to  the 
base  DEF,  as  the  pyramid  ABCG  to  any  solid  which  is  greater 
than  the  pyramid  DEFH.  And  it  has  been  proved,  that  neither 
is  the  base  ABC  to  the  base  DEF,  as  the  pyramid  ABCG  to  any 
solid  which  is  less  than  the  pyramid  DEFH.  Therefore,  as  the 
base  ABC  is  to  the  base  DEF,  so  is  the  pyramid  ABCG  to  the 
pyramid  DEFH.     Wherefore  pyramids,  Sec.  Q.  E.  D. 


PROP.  VI.  THEOR. 

Pyramids  of  the  same  altitude,  which  have  polygons 
for  their  bases,  are  to  one  another  as  their  bases.t 

Let  the  pyramids  which  have  the  polygons  ABCDE,  FGHKL 
for  their  bases,  and  their  vertices  in  the  points  M,  N,  be  of  the 
same  altitude:  as  the  base  ABCDE  to  the  base  FGHKL,  so  is 
the  pyramid  ABCDEM  to  the  pyramid  FGHKLN. 

Divide  the  base  ABCDE  into  the  triangles  ABC,  ACD,  ADE; 
and  the  base  FGHKL  into  the  triangles  FGH,  FHK,  FKL:  and 
upon  the  bases  ABC,  ACD,  ADE  let  there  be  as  many  pyramids 
of  which  the  common  vertex  is  the  point  M,  and  upon  the  remain- 
ing bases  as  many  pyramids  having  their  common  vertex  in  the 


M 


N 


F 


K 


B 


H 


"  This  may  be  explained  the  same  way  as  the  like  at  the  mark  t  in  prop.  2. 
t   See  Note 


212  THE  ELEMENTS  OF  EUCLID.  BOOK  XII, 

point  N:    therefore,  since  the  triangle  ABC  is  to  the  triangle 
FGH,  as  (5.  12.)  the  pyramid  ABCM  to  the  pyramid  FGHN; 
and  the  triangle  ACD   to   the   triangle  FGH,  as  the   pyramid 
ACDM  to  the  pyramid  FGHN;  and  also  the  triangle  ADE  to 
the  triangle  FGH,  as  the  pyramid  ADEM  to  the  pyramid  FGHN5 
as  all  the  first  antecedents  to  their  common  consequent,  so  (2  Cor. 
24.  5.)  are  all  the  other  antecedents  to  their  common  consequent: 
that  is  as  the  base  ABCDE  to  the  base  FGH,  so  is  the  pyramid 
ABCDEM  to  the  pyramid  FGHN:  and,  for  the  same  reason,  as 
the  base  FGHKL  to  the  base  FGH,  so  is  the  pyramid  FGHKLN 
to  the  pyramid  FGHN:  and,  by  inversion,  as  the  base  FGH  to 
the  base  FGHKL;    so  is  the  pyramid  FGHN   to   the   pyramid 
FGHKLN:  then,  because  as  the  base  ABCDE  to  the  base  FGH, 
so  is  the  pyramid  ABCDEM  to  the  pyramid  FGHN;  and  as  the 
base  FGH  to  the  base  FGHKL,  so  is  the  pyramid  FGHN  to  the 
pyramid   FGHKLN;    therefore,  ex  dequali,  (22.  5.)   as  the  base 
ABCDE  to  the  base  FGHKL,  so  the  pyramid  ABCDEM  to  the 
pyramid  FGHKLN.     Therefore,  pyramids,  Sec.     Q.  E.  D. 

PROP.  VH,   THEOR. 

Every  prism  having  a  triangular  base,  may  be  di- 
vided into  three  pyramids  that  have  triangular  bases, 
and  are  equal  to  one  another. 

Let  there  be  a  prism  of  which  the  base  is  the  triangle  ABC, 
and  let  DEF  he  the  triangle  opposite  to  it:  the  prism  ABCDEF 
may  be  divided  into  three  equal  pyramids  having  triangular 
bases. 

Join  BD,  EC,  CD;  and  because  ABED  is  a  parallelogram  of 
which  BD  is  the  diameter,  the  triangle  ABD  is  equal  (34.  I.)  to 
the  triangle  EBD;  therefore  the  pyramid  of  which  the  base  is  the 
triangle  ABD,  and  vertex  the  point  C,  is  equal  (5.  12.)  to  the  py- 
ramid of  which  the  base  is  the  triangle  EBD,  and  vertex  the  point 
C;  but  this  pyramid  is  the  same  with  the  pyramid  the  base  of 
which  is  the  triangle  EBC,  and  vertex  the  point  D;  for  they  are 
contained  by  the  same  planes:  therefore  the  pyramid  of  which 
the  base  is  the  triangle  ABD,  and  vertex  the  point  C,  is  equal  to 
the  pyramid,  the  base  of  which  is  the  triangle  EBC,  and  vertex 
the  point  D:  again,  because  FCBE,  is  a  pa-  F 

rallelogram  of  which  the  diameter  is  CE,  the 
triangle  ECF  is  equal  (34.  1.)  to  the  triangle 
ECB:  therefore  the  pyramid  of  which  the  D 
base  is  the  triangle  ECB,  and  vertex  the  point 
D,  is  equal  to  the  pyramid,  the  base  of  which 
is  the  triangle  ECF,  and  vertex  the  point  D: 
but  the  pyramid  of  which  the  base  is  the  tri- 
angle ECB,  and  vertex  the  point  D,  has  been 
proved  equal  to  the  pyramid  of  which  the  A 
base  is  the  triangle  ABD,  and  vertex  the  point  C.  Therefore  the 
prism  ABCDEF  is  divided  into  three  equal  pyramids  having  tri- 
angular bases,  viz.  into  the  pyramids  ABDC,  EBDC,  ECFD:  'and 


BOOK  XII. 


THE    ELEMENTS  OF  EUCLID. 


213 


because  the  pyramid  of  which  the  base  is  the  triangle  ABD,  and 
vertex  the  point  C,  is  the  same  with  the  pyramid  of  which  the 
base  is  the  triangle  ABC,  and  vertex  the  point  D,  for  they  are 
contained  by  the  same  planesj  and  that  the  pyramid  of  which  the 
base  is  the  triangle  ABD,  and  vertex  the  point  C,  has  been  de- 
monstrated to  be  a  third  part  of  the  prism,  the  base  of  which  is 
the  triangle  ABC,  and  to  which  DEF  is  the  opposite  triangle^ 
therefore  the  pyramid  of  which  the  base  is  the  triangle  ABC,  and 
vertex  the  point  D,  is  the  third  part  of  the  prism  which  has  the 
same  base,  viz.  the  triangle  ABC,  and  DEF  is  the  opposite  trian- 
gle.    Q.  E.  D. 

CoR.  1.  From  this  it  is  manifest,  that  every  pyramid  is  the 
third  part  of  a  prism  which  has  the  same  base,  and  is  of  an  equal 
altitude  with  it^  for  if  the  base  of  the  prism  be  any  other  figure 
than  a  triangle,  it  may  be  divided  into  prisms  having  triangular 
bases. 

CoR.  2.  Prisms  of  equal  altitudes  are  to  one  another  as  their 
bases;  because  the  pyramids  upon  the  same  bases,  and  of  the 
same  altitude,  are  (6.  12.)  to  one  another  as  their  bases. 

PROP.  VIII.  THEOR. 

Similar  pyramids  having  triangular  bases  are  one  to 
another  in  the  triplicate  ratio  of  that  of  their  homolo- 
gous sides. 

Let  the  pyramids  having  the  triangles  ABC,  DEF  for  their 
bases,  and  the  points  G,  H  for  their  vertices,  be  similar,  and  si- 
milarly situated;  the  pyramid  ABCG  has  to  the  pyramid  DEFH, 
the  triplicate  ratio  of  that  which  the  side  BC  has  to  the  homolo- 
gous side  EF. 

Complete  the  parallelograms  ABCM,  GBCN,  ABGK,  and  the 
solid  parallelopiped  BGML  contained  by  these  planes  and  those 


K 


I. 


A 


N 


B 


R 


E 


opposite  to  them:  and,  in  like  manner,  complete  the  solid  paral- 
lelopiped EHPO  contained  by  the  three  parallelograms  DEEP, 
HEFR,  DEHX,  and  those  opposite  to  them:  and,  because  the 
pyramid  ABCG  is  similar  to  the  pyramid  DEFH,  the  angle  ABC 
is  equal  (ll.  def.  1].)  to  the  angle  DEF,  and  the  angle  GBC  to 


214  THE  ELEMENTS  OF  EUCLID.  BOOK  XII. 

the  angle  HEF,  and  ABG  to  DEH:  and  AB  is  (1.  def.  6.)  to  BC 
as  DE  to  EF;  that  is,  the  sides  about  the  equal  angles  are  pro- 
portionals; wherefore  the  parallelogram  BM  is  similar  to  EP:  for 
the  same  reason,  the  parallelogram  BN  is  similar  to  ER,  and  BK 
to  EX;  therefore  the  three  parallelograms  BM,  BN,  BK  are  simi- 
lar to  the  three  EP,  ER,  EX;  but  the  three  BM,  BN,  BK,  are 
equal  and  similar  (24.  11.)  to  the  three  which  are  opposite  to  them, 
and  the  three  EP,  ER,  EX,  equal  and  similar  to  the  three  opposite 
to  them:  wherefore  the  solids  BGML,  EHPO  are  contained  by  the 
same  number  of  similar  planes;  and  their  solid  angles  are  equal 
(B.  1 1.);  and  therefore  the  solid  BGML,  is  similar  (11.  def.  11.)  to 
the  solid  EHPO:  but  similar  solid  parallelopipeds  have  the  tripli- 
cate (33.  1 1.)  ratio  of  thatwhicR  their  homologous  sides  have:  there- 
fore the  solid  BGML  has  to  the  solid  EHPO  the  triplicate  ratio 
of  that  which  the  side  BC  has  to  the  homologous  side  EF:  but  as 
the  solid  BGML  is  to  the  solid  EHPO,  so  is  (15.  5.)  the  pyramid 
ABCG  to  the  pyramid  DEFH;  because  the  pyramids  are  the  sixth 
part  of  the  solids;  since  the  prism,  which  is  the  half  (28.  11.)  of  the 
solid  parallelopiped  is  triple  (7.  12.)  of  the  pyramid.  Wherefore 
likewise  the  pyramid  ABCG  has  to  the  pyramid  DEFH  the  tri- 
plicate ratio  of  that  which  BC  has  to  the  homologous  side  EF. 
Q.  E.  D. 

Cor.  From  this  it  is  evident,  that  similar  pyramids  which  have 
multangular  bases,  are  likewise  to  one  another  in  the  triplicate 
ratio  of  their  homologous  sides:  for  they  may  be  divided  into  simi- 
lar pyramids  having  triangular  bases,  because  the  similar  poly- 
gons, which  are  their  bases,  may  be  divided  into  the  same  number 
of  similar  triangles  homologous  to  the  whole  polygons;  therefore 
as  one  of  the  triangular  pyramids  in  the  first  multangular  pyramid 
is  to  one  of  the  triangular  pyramids  in  the  other,  so  are  all  the 
triangular  pyramids  in  the  first  to  all  the  triangular  pyramids  in 
the  other;  that  is,  so  is  the  first  multangular  pyramid  to  the  other: 
but  one  triangular  pyramid  is  to  its  similar  triangular  pyramid,  in 
the  triplicate  ratio  of  their  homologous  sides;  and  therefore  the 
first  multangular  pyramid  has  to  the  other,  the  triplicate  ratio  of 
that  which  one  of  the  sides  of  the  first  has  to  the  homologous  side 
of  the  other. 

PROP.  IX.   THEOR. 

The  bases  and  altitudes  of  equal  pyramids  having 
triangular  bases  are  reciprocally  proportional:  and  tri- 
angular pyramids  of  which  the  bases  and  altitudes  are 
reciprocally  proportional,  are  equal  to  one  another. 

Let  the  pyrannids  of  which  the  triangles  ABC,  DEF  are  the 
bases,  and  which  have  their  vertices  in  the  points  G,  H,  be  equal 
to  one  another:  the  bases  and  altitudes  of  the  pyramids  ABCG, 
DEFH  are  reciprocally  proportional,  viz.  the  base  ABC  is  to  the 
base  DEF,  as  the  altitude  of  the  pyramid  DEFH  to  the  altitude  of 
the  pyramid  ABCG. 

Complete  the  parallelograms  AC,  AG,  GC,  DF,  DH,  HF,  and 


BOOK  XII. 


THE  ELEMENTS  OF  EUCLID, 


215 


the  solid  parallelopipeds  BGML,  EHPO  contained  by  these  planes 
and  those  opposite  to  them:  and  because  the  pyramid  ABCG  is 
equal  to  the  pyramid  DEFH,  and  that  the  solid  BGML  is  sectuple 
of  the  pyramid  ABCG,  and  the  solid  EHPO  sectuple  of  the 
pyramid  DEFH;  therefore  the  solid  BGML  is  equal  (L  Ax.  5.) 
to  the  solid  EHPO:  but  the  bases  and  aliitudes  of  equal  solid 
parallelopipeds  are  reciprocally  proportional  (34.  11.);  therefore 
as  the  base  BM  to  the  base  EP,  so  is  the  altitude  of  the  solid 

O  R 


N 


K 


c 

y 

\/ 

\ 

M^ 

/' 

c 


-V   ]• 


B 


D 


EHPO  to  the  altitude  of  the  solid  BGML:  but  as  the  base  BM 
to  the  base  EP,  so  is  (15.  5.)  the  triangle  ABC  to  the  triangle 
DEF;  therefore  as  the  triangle  ABC  to  the  triangle  DEF,  so  is 
the  altitude  of  the  solid  EHPO  to  the  altitude  of  the  solid  BGML: 
but  the  altitude  of  the  solid  EHPO  is  the  same  with  the  altitude 
of  the  pyramid  DEFH;  and  the  altitude  of  the  solid  BGML  is  the 
same  with  the  altitude  of  the  pyramid  ABCG:  therefore,  as  the 
base  ABC  to  the  base  DEF,  so  is  the  altitude  of  the  pyramid 
DEFH  to  the  altitude  of  the  pyramid  ABCG:  wherefore  the  bases 
and  altitudes  of  the  pyramids  ABCG,  DEFH  are  reciprocally 
proportional. 

Again,  let  the  bases  and  altitudes  of  the  pyramids  ABCG, 
DEFH  be  reciprocally  proportional,  viz.  the  base  ABC  to  the  base 
DEP',  as  the  altitude  of  the  pyramid  DEFH  to  the  altitude  of  the 
pyramid  ABCG:  the  pyramid  ABCG  is  equal  to  the  pyramid 
DEFH. 

The  same  construction  being  made,  because  as  the  base  ABC  to 
the  base  DEF,  so  is  the  altitude  of  the  pyramid  DEFH  to  the  al- 
titude of  the  pyramid  ABCG:  and  as  the  base  ABC  to  the  base 
DEF,  so  is  the  parallelogram  BM  to  the  parallelogram  EP:  there- 
fore the  parallelogram  BM  is  to  EP,  as  the  altitude  of  the  pyramid 
DEFH  to  the  altitude  of  the  pyramid  ABCG:  but  the  altitude  of 
the  pyramid  DEFH  is  the  same  with  the  altitude  of  the  solid  pa- 
rallelopiped  EHPO:  and  the  altitude  of  the  pyramid  ABCG  is  the 
same  with  the  altitude  of  the  solid  parallelopiped  BGML:  as, 
therefore,  the  base  BM  to  the  base  EP,  so  is  the  altitude  of  the 
solid  parallelopiped  EHPO  to  the  altitude  of  the  solid  parallelo- 
piped BGML.  But  solid  parallelopipeds  having  their  bases  and 
altitudes  reciprocally  proportional,  are  equal  (34.  11.)  to  one  ano- 
ther.    Therefore  the  solid  parallelepiped  BGML  is  equal  to  the 


216  THE  ELEMENTS  OF  EUCLID.  BOOK  XII* 

solid  parallelopiped  EHPO.  And  the  pyramid  ABCG  is  the 
sixth  part  of  the  solid  BGML,  and  the  pyramid  DEFH  is  the 
sixth  part  of  the  solid  EPHO.  Therefore  the  pyramid  ABCD  is 
equal  to  the  pyramid  DEFH.    Therefore  the  bases,  Sec.  Q.  E.  D. 

PROP.  X.  THEOR. 

Every  cone  is  the  third  part  of  a  cyhnder  which  has 
the  same  base,  and  is  of  an  equal  altitude  with  it. 

Let  a  cone  have  the  same  base  with  a  cylinder,  viz.  the  circle 
ABCDj  and  the  same  altitude.  The  cone  is  the  third  part  of  the 
cylinder^  that  is,  the  cylinder  is  triple  of  the  cone. 

If  the  cylinder  be  not  triple  of  the  cone,  it  must  either  be  greater 
than  the  triple,  or  less  than  it.  First,  let  it  be  greater  than  the 
triple:  and  describe  the  square  ABCD  in  the  circle;  this  square  is 
greater  than  the  half  of  the  circle  ABCD:*  Upon  the  square 
ABCD  erect  a  prism  of  the  same  altitude  with  the  cylinder;  this 
prism  is  greater  than  half  of  the  cylinder:  because  if  a  square  be 
described  about  the  circle,  and  a  prism  erected  upon  the  square, 
of  the  same  altitude  with  the  cylinder,  the  inscribed  square  is  half 
of  that  circumscribed;  and  upon  these  square  bases  are  erected 
solid  parallelopipeds,  viz.  the  prisms  of  the  same  altitude;  there- 
fore the  prism  upon  the  square  ABCD  is  half  of  the  prism  upon 
the  square  described  about  the  circle:  because  they  are  to  one 
another  as  their  bases  (32.  11.);  and  tlie  cylinder  is  less  than  the 
prisin  upon  the  square  described  about  the  circle  ABCD:  there- 
fore the  prism  upon  the  square  ABCD  of  the  same  altitude  with 
the  cylinder,  is  greater  than  half  of  the  cylinder.  Bisect  the  cir- 
cumferences AB,  BC,  CD,  DA  in  the  points  E,  F,  G,  H,  and  join 
AE,  EB,  BF,  FC,  CG,  GD,  DH,  HA:  then  each  of  the  triangles 
AEB,  BFC,  CGD,  DHA  is  greater  than  half  of  the  segment  of  the 
circle  in  which  it  stands,  as  was  shown  in  prop.  2.  of  this  book. 
Erect  prisms  upon  each  of  these  triangles,  of  the  same  altitude 
with  the  cylinder;  each  of  these  prisms  A 
is  greater  than  half  of  the  segment  of 
the  cylinder  in  which  it  is;  because  if 
through  the  points  E,  F,  G,  H,  parallels 
be  drawn  to  AB,  BC,  CD,  DA,  and  pa- 
rallelograms be  completed  upon  the      B  ^^ ^  D 

same  AB,  BC,  CD,  DA,  and  solid  pa- 
rallelopipeds be.  erected  upon  the  pa- 
rallelograms; the  prisms  upon  the  tri- 
angles AEB,  BFC,  CGD,  DHA  are  the 
halves  of  the  solid  parallelopipeds  (2.  C 

Cor.  7.  12.).  And  the  segments  of  the  cylinder  which  are  upon 
the  segments  of  the  circle  cut  off  by  AB,  BC,  CD,  DA,  are  less 
than  the  solid  parallelopipeds  which  contain  them.  Therefore  the 
prisms  upon  the  triangles  AEB,  BFC,  CGD,  DHA,  are  greater 
than  half  of  the  segments  of  the  cylinder  in  which  they  are;  there - 

*  As  was  shown  in  prop.  2.  of  this  book. 


iiOOK   XII. 


THE  ELEMENTS  OF  EUCLID. 


217 


fore  if  each  of  the  circumferences  be  divided  into  two  equal 
parts,  and  straight  lines  be  drawn  from  the  points  of  division  to 
the  extremities  of  the  circumferences,  and  upon  the  triangles 
thus  made,  prisms  be  erected  of  the  same  altitude  with  the  cy- 
linder, and  so  on,  there  must  at  length  remain  some  segments 
of  the  cylinder  which  together  are  less  (Lem.)  than  the  ex- 
cess of  the  cylinder  above  the  triple  of  the  cone.  Let  them 
be  those  upon  the  segments  of  the  circle  AE,  EB,  BF,  FC, 
CG,  GD,  DH,  HA.  Therefore  the  rest  of  the  cylinder,  that  is, 
the  prism  of  which  the  base  is  the  polygon  AEBFCGDH,  and 
of  which  the  altitude  is  the  same  with  that  of  the  cylinder,  is 
greater  than  the  trii)Ie  of  the  cone:  but  this  prism  is  triple  (1. 
Cor.  7.  12.)  of  the  pyramid  upon  the  same  base,  of  which  the  ver- 
tex is  the  same  with  the  vertex  of  the  conej  therefore  the  pyra- 
mid upon  the  base  AEBFCGDH,  having  the  same  vertex  with 
the  cone,  is  greater  than  the  cone,  of  which  the  base  is  the  circle 
ABCD:  but  it  is  also  less,  for  the  pyramid  is  contained  within 
the  cone;  wliich  is  impossible.  Nor  can  the  cylinder  be  less  than 
the  triple  of  the  cone.  Let  it  be  less,  if  possible:  therefore,  in- 
versely, the  cone  is  greater  than  the  third  part  of  the  cylinder. 
In  the  circle  ABCD  describe  a  square;  this  square  is  greater 
than  the  half  of  the  circle:  and  upon  the  square  ABCD  erect  a 
pyramid  having  the  same  vertex  with  the  cone:  this  pyramid  is 
greater  than  the  half  of  the  cone;  because,  as  was  before  demon- 
strated, if  a  square  be  described  about  H 
^he  circle,  the  square  ABCD  is  the 
half  of  it;  and  if,  upon  these  squares, 
there  be  erected  solid  parallelopipeds 
of  the  same  altitudes  with  the  cone, 
which  are  also  prisms,  the  prism  upon 
the  square  ABCD  shall  be  the  half  of 
that  which  is  upon  the  square  de- 
scribed about  the  circle;  for  they  are 
to  one  another  as  their  bases  (32.  1 1.); 
as  are  also  the  third  parts  of  them; 
therefore  the  pyramid,  the  base  of  F 
which  is  the  square  ABCD,  is  half  of  the  pyramid  upon  the  square 
described  about  the  circle:  but  this  last  pyramid  is  greater  than 
the  cone  which  it  contains;  therefore  the  pyramid  upon  the  square 
ABCD,  having  the  same  vertex  with  the  cone,  is  greater  than  the 
half  of  the  cone.  Bisect  the  circumferences  AB,  BC,  CD,  DA  in 
the  points  E,  F,  G,  H,  and  join  AE,  EB,  BF,  FC,  CG,  GD,  DH, 
HA:  therefore  each  of  the  triangles  AEB,  BFC,  CGD,  DHA  is 
greater  than  half  of  the  segment  of  the  circle  in  which  it  is:  upon 
each  of  these  triangles  erect  pyramids  having  the  same  vertex 
with  the  cone.  Therefore  each  of  these  pyramids  is  greater  than 
the  half  of  the  segment  of  the  cone  in  which  it  is,  as  before  was 
demonstrated  of  the  prisms  and  segments  of  the  cylinder:  and 
thus  dividing  each  of  the  circumferences  into  two  equal  parts, 
and  joining  the  points  of  division  and  their  extremities  by  straight 
lines,  and  upon  the  triangles  erecting  pyramids  having  their  ver- 
tices the  same  with  that  of  the  cone,  and  so  on,  there  must  at 

2  E 


218 


THE  ELEMENTS  OF  EUCLID. 


BOOK  Xlly^ 


length  remain  some  segments  of  the  cone,  which  together  shall 
be  less  than  the  excess  of  the  cone  above  the  third  part  of  the  cy- 
linder. Let  these  be  the  segments  upon  AE,  EB,  BF,  FC,  CG^ 
GD,  DH,  HA.  Therefore  the  rest  of  the  cone,  that  is,  the  pyra- 
mid, of  which  the  base  is  the  polygon  H 
AEBFCGDH,  and  of  which  the  vertex, 
is  the  same  with  that  of  the  cone,  is 
greater  than  the  third  part  of  the  cy- 
linder. But  this  pyramid  is  the  third 
part  of  the  prism  upon  the  same  base  E 
AEBFCGDH,  and  of  the  same  alti- 
tude with  the  cylinder.  Therefore 
this  prism  is  greater  than  the  cylinder 
of  which  the  base  is  the  circle  ABCD. 
But  it  is  also  less;  for  it  is  contained 
within  the  cylinder;  which  is  impos- 
sible. Therefore  the  cylinder  is  not  less  than  the  triple  of  the 
cone.  And  it  has  been  demonstrated  that  neither  is  it  greater 
than  the  triple.  Therefore  the  cylinder  is  triple  of  the  cone,  or 
the  cone  is  the  third  part  of  the  cylinder.  Wherefore  every  concj 
Sec.     Q.  E.  D. 

PROP.  XI.  THEOR. 

Cones  and  cylinders  of  the  same  altitude,  are  to  one 
another  as  their  bases.* 

Let  the  cones  and  cylinders,  of  which  the  bases  are  the  circles 
ABCD,  EFGH,  and  the  axes  KL,  MN,  and  AC,  EG  the  diame- 
ters of  their  bases,  be  of  the  same  altitude.  As  the  circle  ABCD 
to  the  circle  EFGH,  so  is  the  cone  AL  to  the  cone  EN. 

If  it  be  not  so,  let  the  circle  ABCD  be  to  the  circle  EFGH,  as 
the  cone  AL  to  some  solid  either  less  than  the  cone  EN,  or  greater 
than  it.  First,  let  it  be  to  a  solid  less  than  EN,  viz.  to  the  solid 
X;  and  let  Z  be  the  solid  which  is  equal  to  the  excess  of  the  cone 
EN  above  the  solid  X;  therefore  the  cone  EN  is  equal  to  the  so- 
lids X,  Z  together.  In  the  circle  EFGH  describe  the  square 
EFGH,  therefore  this  square  is  greater  than  the  half  of  the  circle: 
upon  the  square  EFGH  erect  a  pyramid  of  the  same  altitude  with 
the  cone;  this  pyramid  is  greater  than  half  of  the  cone.  For,  if 
a  square  be  described  about  the  circle,  and  a  pyramid  be  erected 
upon  it,  having  the  same  vertex  with  the  cone,t  the  pyramid  in- 
scribed in  the  cone  is  half  of  the  pyramid  circumscribed  about  it, 
because  they  are  to  one  another  as  their  bases  (6.  12.):  but  the 
cone  is  less  than  the  circumscribed  pyramid;  therefore  the  pyra- 
mid of  which  the  base  is  the  square  EFGH,  and  its  vertex  the 
same  with  that  of  the  cone,  is  greater  than  half  of  the  cone:  di- 
vide the  circumferences  EF,  FG,  GH,  HE,  each  into  two  equal 

*  See  Note. 

t  Vertex  is  put  ip  place  of  altitude,  which  is  in  the  Greek,  because  the  pyra- 
mid, in  what  follows,  is  supposed  to  be  circumscribed  about  the  cone,  and  so 
must  have  the  same  vertex.  And  the  same  change  is  made  in  some  places  fol- 
lowing. 


.  BOOK  XII. 


THE  ELEMENTS  OF  EUCLID. 


219 


parts  in  the  points  O,  P,  R,  S,  and  join  EO,  OF,  FP,  PG,  GR, 
RH,  HS,  SE:  therefore  each  of  the  triangles  EOF,  FPG,  GRH, 
HSE  is  greater  than  half  of  the  segment  of  the  circle  in  which  it 


C       E 


G 


K 


x 


Iv 


X 


nx 


K-z\ 


is:  upon  each  of  these  triangles  erect  a  pyramid  having  the  same 
vertex  with  the  cone;  each  of  these  pyramids  is  greater  than  the 
half  of  the  segment  of  the  cone  in  which  it  is:  and  thus  dividing 
each  of  these  circumferences  into  two  equal  parts,  and  from  the 
points  of  division  drawing  straight  lines  to  the  extremities  of  the 
circumferences,  and  upon  each  of  the  triangles  thus  made  erect- 
ing pyramids,  having  the  same  vertex  with  the  cone,  and  so  on, 
there  must  at  length  remain  some  segments  of  the  cone  which  are 
together  less  (Lem.  1.)  than  the  solid  Z:  let  these  be  the  segments 
upon  EO,  OF,  FP,  PG,  GR,  RH,  HS,  SE:  therefore  the  remain- 
der of  the  cone,  viz.  the  pyramid  of  which  the  base  is  the  polygon 
EOFPGRHS,  and  its  vertex  the  same  with  that  of  the  cone,  is 
greater  than  the  solid  X:  in  the  circle  A  BCD  describe  the  poly- 
gon ATBYCVDQ  similar  to  the  polygon  EOFPGRHS,  and  upon 
it  erect  a  pyramid  having  the  same  vertex  with  the  cone  AL:  and 
because  as  the  square  of  AC  is  to  the  square  of  EG,  so  (1.  12.)  is 
the  polygon  ATBYCVDQ  to  the  polygon  EOFPGRHS,-  and  as 
the  square  of  AC  to  the  square  of  EG,  so  is  (2.  12.)  the  circle 
ABCD  to  the  circle  EFGH;  therefore  the  circle  ABCD  (11.  5.) 
is  to  the  circle  EFGH,  as  the  polygon  ATBYCVDQ  to  the  poly- 
gon EOFPGRHS:  but  as  the  circle  ABCD  to  the  circle  EFGH, 
so  is  the  cone  AL  to  the  solid  X,  and  as  the  polygon  ATBYCVDQ 
to  the  polygon  EOFPGRHS,  so  is  (6.  12.)  the  pyramid  of  which 
the  base  is  the  first  of  these  polygons,  and  vertex  L,  to  the  pyra- 
ssiid  of  v/hich  the  base  is  the  other  polygon,  and  its  vertex  N: 


220 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XII. 


therefore,  as  the  cone  AL  to  the  solid  X,  so  is  the  pyramid  of 
which  the  base  is  the  polygon  ATBYCVDQ,  and  vertex  L,  to  the 
pyramid  the  base  of  which  is  the  polygon  EOFPGRHS,  and  ver- 
tex N;  but  the  cone  AL  is  greater  than  the  pyramid  contained  in 


C       K 


B 


V 


it;  therefore  the  solid  X  is  greater  (14.  5.)  than  the  pyramid  in 
the  cone  EN5  but  it  is  less,  as  was  shown,  which  is  absurd:  there- 
fore the  circle  ABCD  is  not  to  the  circle  EFGH,  as  the  cone  AL 
to  any  solid  which  is  less  than  the  cone  EN.  In  the  same  man- 
ner it  may  be  demonstrated  that  the  circle  EFGH  is  not  to  the 
circle  ABCD,  as  the  cone  EN  to  any  solid  less  than  the  cone  AL. 
Nor  can  the  circle  ABCD  be  to  the  circle  EFGH,  as  the  cone  AL 
to  any  solid  greater  than  the  cone  EN:  for,  if  it  be  possible,  let  it 
be  so  to  the  solid  I,  which  is  greater  than  the  cone  EN:  there- 
fore, by  inversion,  as  the  circle  EFGH  to  the  circle  ABCD,  so  is 
the  solid  I  to  the  cone  AL:  but  as  the  solid  I  to  the  cone  AL,  so 
is  the  cone  EN  to  some  solid  which  must  be  less  (14.  5.)  than  the 
cone  AL,  because  the  solid  I  is  greater  than  the  cone  EN:  there- 
fore as  the  circle  EFGH  is  to  the  circle  ABCD,  so  is  the  cone 
EN  to  a  solid  less  than  the  cone  AL,  which  was  shown  to  be  im- 
possible; therefore  the  circle  ABCD  is  not  to  the  circle  EFGH, 
as  the  cone  AL  is  to  any  solid  greater  than  the  cone  EN:  and  it 
has  been  demonstrated  that  neither  is  the  circle  ABCD  to  the  cir- 
cle EFGH,  as  the  cone  AL  to  any  solid  less  than  the  cone  EN: 
therefore  the  circle  ABCD  is  to  the  circle  EFGH,  as  the  cone 
AL  to  the  cone  EN:  but  as  the  cone  is  to  the  cone,  so  (15.  5.)  is 
the  cylinder  to  the  cylinder,  because  the  cylinders  are  triple  (10. 


nooK  XII. 


THE  ELEMENTS  OF  EUCLID. 


221 


12.)  of  the  cone,  each  to  each.  Therefore,  as  the  circle  ABCD  to 
the  circle  EFGH,  so  are  the  cylinders  upon  them  of  the  same  al- 
titude. Wherefore  cones  and  cylinders  of  the  same  altitude  are 
to  one  another  as  their  bases.     Q.  E.  D. 

PROP.  XII.  THEOR. 

Similar  cones  and  cylinders  have  to  one  another  the 
tripHcate  ratio  of  that  which  the  diameters  of  their 
bases  have.* 

Let  the  cones  and  cylinders  of  which  the  bases  arc  the  circles 
ABCD,  EFGH,  and  the  diameters  of  the  bases,  AC,  EG,  and  KL, 
MN  the  axes  of  the  cones  or  cylinders,  be  similar:  the  cone  of 
which  the  base  is  the  circle  ABCD,  and  vertex  the  point  L,  has 
to  the  cone  of  which  the  base  is  the  circle  EFGH,  and  vertex  N, 
the  triplicate  ratio  of  that  which  AC  has  to  EG. 

For,  if  the  cone  ABCDL  has  not  to  the  cone  EFGHN  the  tri- 
plicate ratio  of  that  v/hich  AC  has  to  EG,  the  cone  ABCDL  shall 
have  the  triplicate  of  that  ratio  to  some  solid  which  is  less  or 
greater  than  the  cone  EFGHN.  First,  let  it  have  it  to  a  less,  viz. 
to  the  solid  X.  Make  the  same  construction  as  in  the  preceding 
proposition,  and  it  may  be  demonstrated  the  very  same  way  as  in 
that  proposition,  that  the  pyramid  of  which  the  base  is  the  poly- 


See  Note 


222  THE  ELEMENTS  OF  EUCLID.  BOOK  XII, 

got!  EOFPGRHS,  and  vertex  N,  is  greater  than  the  solid  X.    De- 
scribe also  in  the  circle  ABCD  the  polygon  ATBYCVDQ  simi- 
lar to  the  polyg-on  EOFPGRHS,  upon  which  erect  a  pyramid 
having  the  same  vertex  with  the  cone;  and  let  LAQ  be  one  of  the 
triangles  containing  the  pyramid  upon  the  polygon  ATBYCVDQ 
the  vertex  of  which  is  L;  and  let  NES  be  one  of  the  triangles  con- 
taining the  pyramid  upon  the  polygon  EOFPGRHS  of  which  the 
vertex  is  N;  and  join  KQ,  MS:  because  then  the  cone  ABCDL 
is  similar  to  the  cone  EFGHN,  AC  is  (24.  def.  1 1.)  to  EG,  as  the 
axis  KL  to  the  axis  MN;  and  as  AC  to  EG,  so  (15.  5.)  is  AK  to 
EM;  therefore  as  AK  to  EM,  so  is  KL  to  MN;  and,  alternately, 
AK  to  KL,  as  EM  to  MN:  and  the  right  angles  AKL,  EMN  are 
equal;  therefore  the  sides  about  these  equal  angles  being  propor- 
tionals, the  triangle  AKL  is  similar  (6.  6.)  to  the  triangle  EMN. 
Again,  because  AK  is  to  KQ,  as  EM  to  MS,  and  that  these  sides 
are  about  equal  angles  AKQ,  EMS,  because  these  angles  are,  each 
of  them,  the  same  part  of  four  right  angles  at  the  centres  K,  M; 
therefore  the  triangle  AKQ  is  similar  (6.  6.)  to  the  triangle  EMS: 
and  because  it  has  been  shown,  that  as  AK  to  KL,  so  is  EM  to 
MN,  and  that  AK  is  equal  to  KQ,  and  EM  to  MS,  as  QK  to  KL, 
so  is  SM  to  MN,  and  therefore  the  sides  about  the  right  angles 
QKL,  SMN  being  proportionals,  the  triangle  LKQ  is  similar  to 
the  triangle  NMS:  and  because  of  the  similarity  of  the  triangles 
AKL,  EMN,  as  LA  is  to  AK,  so  is  NE  to  EM:  and  by  the  simi- 
larity of  the  triangles  AKQ,  EMS,  as  KA  to  AQ,  so  ME  to  ES; 
ex  dequali,  (22.  5.)  LA  is  to  AQ  as  NE  to  ES.     Again,  because  of 
the  similarity  of  the  triangles  LQK,  NSM;  as  LQ  to  QK,  so  NS 
to  SM;  and  from  the  similarity  of  the  triangles  KAQ,  MES,  as 
KQ  to  QA,  so  MS  to  SE;  ex  asquali,  (22.  5.)  LQ  is  to  QA,  as  NS 
to  SE:  and  it  was  proved  that  QA  is  to  AL  as  SE  to  EN,  there- 
fore, again,  ex  sequali,  as  QL  to  LA,  so  is  SN  to  NE;  wherefore 
the  triangles  LQA,  NSE,  having  the  sides  about  all  their  angles 
proportionals,  are  equiangular  (5.  6.)  and  similar  to  one  another: 
and  therefore  the  pyramid  of  which  the  base  is  the  triangle  AKQ 
and  vertex  L,  is  similar  to  the  pyramid  the  base  of  which  is  the 
triangle  EMS,  and  vertex  N,  because  their  solid  angles  are  equal 
(B.  11.)  to  one  another,  and  they  are  contained  by  the  same  num- 
ber of  similar  planes:  but  similar  pyramids  which  have  triangu- 
lar bases  have  to  one  another  the  triplicate  (8.  12.)  ratio  of  that 
which  their  homologous  sides  have;  therefore  the  pyramid  AKQL 
has  to  the  pyramid  EMSN  the  triplicate  ratio  of  that  which  AK 
has  to  EM.     In  the  same  manner,  if  straight  lines  be  drawn  from 
the  points  D,  V,  C,  Y,  B,  T  to  K,  and  from  the  points  H,  R,  G, 
P,  F,  O  to  M,  and  pyramids  be  erected  upon  the  triangles  having 
the  same  vertices  with  the  cones,  it  may  be  demonstrated  that 
each  pyramid  in  the  first  cone  has  to  each  in  the  other,  taking 
them  in  the  same  order,  the  triplicate  ratio  of  that  which  the  side 
AK  has  to  the  side  EM;  that  is,  which  AC  has  to  EG:  but  as 
one  antecedent  to  its  consequent,  so  are  all  the  antecedents  to  all 
the  consequents  (12.  5.);  therefore  as  the  pyramid  AKQL  to  the 
pyramid  EMSN,  so  is  the  whole  pyramid  the  base  of  which  is 
the  polygon  DQATBYCV,  and  vertex  L,  to  the  whole  pyramid 


BOOK  XII. 


THE  ELEMENTS  OF  EUCLID. 


22; 


of  which  the  base  is  the  polygon  HSEOFPGR,  and  vertex  N. 
Wherefore  also  the  first  of  these  two  last  named  pyramids  has  to 
the  other  the  triplicate  ratio  of  that  which  AC  has  to  EG.  But 
by  the  hypothesis,  the  cone  of  which  the  base  is  the  circle  ABCD, 
and  vertex  L,  has  to  the  solid  X,  the  triplicate  ratio  of  that  which 
AC  has  to  EG:  therefore  as  the  cone  of  which  the  base  is  the 
circle  ABCD,  and  vertex  L,  is  to  the  solid  X,  so  is  the  pyramid 
the  base  of  which  is  the  polygon  DQATBYCV,  and  vertex  L,  to 
the  pyramid  the  base  of  which  is  the  polygon  HSEOFPGR,  and 
vertex  N:  but  the  said  cone  is  greater  than  the  pyramid  contained 
in  it,  therefore  the  solid  X  is  greater  (14.  5.)  than  the  pyramid, 
the  base  of  which  is  the  polygon  HSEOFPGR,  and  vertex  N; 
but  it  is  also  less;  which  is  impossible;  therefore  the  cone,  of 
which  the  base  is  the  circle  ABCD,  and  vertex  L,  has  not  to  any 


A 


B 


solid  which  is  less  than  the  cone  of  which  the  base  is  the  circle 
EFGH  and  vertex  N,  the  triplicate  ratio  of  that  which  AC  has  to 
EG.  In  the  same  manner  it  may  be  demonstrated  that  neither 
has  the  cone  EFGHN  to  any  solid  which  is  less  than  the  cone 
ABCDL,  the  triplicate  ratio  of  that  which  EG  has  to  AC.  Nor 
can  the  cone  ABCDL  have  to  any  solid  which  is  greater  than  the 
cone  EFGHN,  the  triplicate  ratio  of  that  which  AC  has  to  EG: 
for,  if  it  be  possible,  let  it  have  it  to  a  greater,  viz.  to  the  solid  Z: 
therefore,  inversely,  the  solid  Z  has  to  the  cone  ABCDL,  the  tri- 
plicate ratio  of  that  which  EG  has  to  AC:  but  as  the  solid  Z  is 


224 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XII. 


to  the  cone  ABCDL,  so  is  the  cone  EFGHN  to  some  solid,  which 
must  be  less  (14.  5.)  than  the  cone  ABCDL,  because  the  solid  Z 
is  greater  than  the  cone  EFGHN:  therefore  the  cone  EFGHN 
has  to  a  solid  which  is  less  than  the  cone  ABCDL,  the  triplicate 
ratio  of  that  which  EG  has  to  AC,  which  was  demonstrated  to 
be  impossible:  therefore  the  cone  ABCDL  has  not  to  any  solid 
greater  than  the  cone  EFGHN,  the  triplicate  ratio  of  that  which 
AC  has  to  EG;  and  it  was  demonstrated  that  it  could  not  have 
that  ratio  to  any  solid  less  than  the  cone  EFGHN:  therefore  the 
cone  ABCDL  has  to  the  cone  EFGHN  the  triplicate  ratio  of  that 
which  AC  has  to  EG:  but  as  the  cone  is  to  the  cone,  so  (15.  5.)  the 
cylinder  to  the  cylinder;  for  every  cone  is  the  third  part  of  the 
cylinder  upon  the  same  base,  and  of  the  same  altitude:  therefore 
also  the  cylinder  has  to  the  cylinder  the  triplicate  ratio  of  that 
which  AC  has  to  EG.     Wherefore  similar  cones,  Sec.     Q.  E.  D. 


PROP.  XHL  THEOR. 

If  a  cylinder  be  cut  by  a  plane,  parallel  to  its  oppo- 
site planes,  or  bases,  it  divides  the  cylinder  into  two 
cylinders,  one  of  which  is  to  the  other  as  the  axis  of 
the  first  to  the  axis  of  the  other. "^ 

Let  the  cylinder  AD  be  cut  by  the  plane 
GH,  parallel  to  the  opposite  planes  AB, 
CD,  meeting  the  axis  EF  in  the  point  K, 
and  let  the  line  GH  be  the  common  sec- 
tion of  the  plane  GH  and  the  surface  of  the 
cylinder  AD:  let  AEFC  be  the  parallelo- 
gram, in  any  position  of  it,  by  the  revolu- 
tion of  which  about  the  straight  line  EF  the 
cylinder  AD  is  described;  and  let  GK  be 
the  common  section  of  the  plane  GH,  and 
the  plane  AEFC:  and  because  the  parallel 
planes  AB,  GH  are  cut  by  the  plane  AEKG, 
AE,  KG,  their  common  sections  with  it, 
are  parallel  (16.  11.):  wherefore  AK  is  a 
parallelogram,  and  GK  equal  to  EA  the 
straight  line  from  the  centre  of  the  circle 
AB:  for  the  same  reason  each  of  the 
straight  lines  drawn  from  the  point  K  to 
the  line  GH  may  be  proved  to  be  equal  to 
those  which  are  drawn  from  the  centre  of 

the  circle  AB  to  its  circumference,  and  are  

therefore  all  equal  to  one  another.  Therefore  the  line  GH  is  the 
circumference  of  a  circle,  (15.  def.  1.)  of  which  the  centre  is  the 
point  K:  therefore  the  plane  GH  divides  the  cylinder  AD  into  the 
cylinders  AH,  GD;  for  they  are  the  same  which  would  be  de- 
scribed by  the  revolution  of  the  parallelograms  AK,  GF,  about 

'  See  Note. 


BOOK  XII. 


THE  ELEMENTS  OF  EUCLID. 


225 


B 


the  straight  lines  EK,  KF:  and  it  is  to  be  shown,  that  the  cylin- 
der AH  is  to  the  cylinder  HC,  as  the  axis  EK  to  the  axis  KF. 

Produce  the  axis  EF  both  ways^  and  take  any  number  of 
straight  lines  EN,  NL,  each  equal  to  EK^  and  any  number  FX, 
XM  each  equal  to  FK;  and  let  planes  pa- 
rallel to  AB,  CD  pass  through  the  points 
L,  N,  X,  M;  therefore  the  common  sec- 
tions of  these  planes  with  the  cylinder  pro- 
duced are  circles  the  centres  of  which  are 
the  points  L,  N,  X,  M,  as  was  proved  of 
the  plane  GH;  and  these  planes  cut  off'  the 
cylinders  PR,  RB,  DT,  TQ:  and  because 
the  axes  LN,  NE,  EK  are  all  equal,  there- 
fore the  cylinders  PR,  RB,  BG  are  (11.13.) 
to  one  another  as  their  bases:  but  their 
bases  are  equal,  and  therefore  the  cylinders 
PR,  RB,  BG  are  equal:  and  because  the 
axes  LN,  NE,  EK  are  equal  to  one  another, 
as  also  the  cylinders  PR,  RB,  BG,  and  that 
there  are  as  many  axes  as  cylinders;  there- 
fore, whatever  multiple  the  axis  KL  is  of 
the  axis  KE,  the  same  multiple  is  the  cy- 
linder PG  of  the  cylinder  GB:  for  the  same 
reason,  whatever  multiple  the  axis  MK  is 
of  the  axis  KF,  the  same  multiple  is  the 
cylinder  QG  of  the  cylinder  GD:  and  if  the  axis  KL  be  equal  to 
the  axis  KM,  the  cylinder  PG  is  equal  to  the  cylinder  GQ;  and  if 
the  axis  KL  be  greater  than  the  axis  KM,  the  cylinder  PG  is 
greater  than  the  cylinder  QG;  and  if  less,  less:  since  therefore 
there  are  four  magnitudes,  viz.  the  axes  EK,  KF,  and  the  cylin- 
ders BG,  GD,  and  that  of  the  axis  EK  and  cylinder  BG,  there  has 
been  taken  any  equimultiples  whatever,  viz.  the  axis  KL  and  cy- 
linder PG;  and  of  the  axis  KF  and  cylinder  GD,  any  equimulti^ 
pies  whatever,  viz.  the  axis  KM  and  cylinder  GQ:  and  it  has  been 
demonstrated,  if  the  axis  KL  be  greater  than  the  axis  KM,  the  cy- 
linder PG  is  greater  than  the  cylinder  GQ;  and  if  equal,  equal; 
and  if  less,  less:  therefore  (5.  def.  5.)  the  axis  EK  is  to  the  axis 
FK,  as  the  cylinder  BG  to  the  cylinder  GD.  Wherefore,  if  a  cy- 
linder, Sec.     Q.  E.  D. 


H 


D 


PROP.  XIV.  THEOR. 

Cones  and  cylinders  upon  equal  bases  are  to  one 
another  as  their  altitudes. 

Let  the  cylinders  EB,  FD  be  upon  the  equal  bases  AB,  CD: 
as  the  cylinder  EB  to  the  cylinder  FD,  so  is  the  axis  GH  to  the 
axis  KL. 

Produce  the  axis  KL  to  the  point  N,  and  make  LN  equal  to 

the  axis  GH,  and  let  CM  be  a  cylinder  of  which  the  base  is  CD, 

and  axis  LN:  and  because  the  cylinders  EB,  CM,  have  the  same 

altitude,  they  are  to  one  another  as  their  bases:  (1 1.  12.)  but  their 

2F 


226 


THE  ELEMENTS  OF  iiUCLlD. 


BOOK  XII. 


bases  are  equal:  therefore  also  the  cylinders  EB,  CM  are  equal; 
And  because  the  cylinder  FM 
is  cut  by  the  plane  CD  parallel 
to  its  opposite  planes,  as  the  cy- 
linder CM  to  the  cylinder  FD, 
so  is  (is.  12.)  the  axis  LN  to 
the  axis  KL.  But  the  cylinder 
CM  is  equal  to  the  cylinder  EB, 
and  the  axis  LN  to  the  axis  GH: 
therefore  as  the  cylinder  EB  to 
the  cylinder  FD,  so  is  the  axis 
GH  to  the  axis  KL:  and  as  the 
cylinder  EB  to  the  cylinder  FD, 
so  is  (15.  5.)  the  cone  ABG  to 


A 


the  cone  CDK,  because  the  cylinders  are  triple  (10.  12.)  of  the 
cones:  therefore  also  the  axis  GH  is  to  the  axis  KL,  as  the  cone 
ABG  to  the  cone  CDK,  and  the  cylinder  EB  to  the  cylinder  FD. 
Wherefore  cones,  &c.     Q.  E.  D. 

PROP.  XV.  THEOR. 

The  bases  and  altitudes  of  equal  cones  and  cylinders 
are  reciprocally  proportional;  and,  if  the  bases  and  al- 
titudes be  reciprocally  proportional,  the  cones  and  cy- 
linders are  equal  to  one  another."^ 

Let  the  circles  ABCD,  EFGH,  the  diameters  of  which  are  AC, 
EG,  be  the  bases,  and  KL,  MN  the  axes,  as  also  the  altitudes  of 
ecjual  cones  and  cylinders;  and  let  ALC,  ENG  be  the  cones,  and 
AX,  EO  the  cylinders:  the  bases  and  altitudes  of  the  cylinders 
AX,  EO  are  reciprocally  proportional;  that  is,  as  the  base  ABCD 
to  the  base  EFGH,  so  is  the  altitude  MN  to  the  altitude  KL. 

Either  the  altitude  MN  is  equal  to  the  altitude  KL,  or  these 
altitudes  are  not  equal.  First,  let  them  be  equal:  and  the  cylin- 
ders AX,  EO,  being  also  equal,  and  cones  and  cylinders  of  the 
same  altitude  being  to  one  another  as  their  bases,  (U.  12.)  there- 
fore the  base  ABCD  is  equal  (A.  5.)  to  the  base  EFGH;  and  as 
the  base  ABCD  is  to  the  base  EFGH,  so  is  the  altitude  MN  to 
the  altitude  KL.  But  let  the 
altitudes  KL,  MN  be  unequal, 
and  MN  the  greater  of  the  two, 
and  from  MN  take  MP  equal 
to  KL,  and,  through  the  point 
P,  cut  the  cylinder  EO  by  the 
plane  TYS,  parallel  to  the  op- 
posite planes  of  the  circles 
EFGH,  RO;  therefore  the  com- 
mon section  of  the  plane  TYS 
and  the  cylinder  EO  is  a  cir- 
cle, and  consequently  ES  is  a 


A 


See  Note, 


BOOK  XII.  THE  liLEMENTS  OF    EUCLID.  227 

cylinder,  the  base  of  which  is  the  circle  EFHG,  and  altitude  MP: 
and  because  the  cylinder  AX  is  equal  to  the  cylinder  EO,  as  AX 
is  to  the  cylinder  ES,  so  (7.  5.)  is  the  cylinder  EO  to  the  same  ES. 
But  as  the  cylinder  AX  to  the  cylinder  ES,  so  (11.  12.)  is  the  base 
ABCD  to  the  base  EFGHj  for  the  cylinders  AX,  ES  are  of  the 
same  altitude;  and  as  the  cylinder  EO  to  the  cylinder  ES,  so  (13. 
12.)  is  the  altitude  MN  to  the  altitude  MP,  because  the  cylinder 
EO  is  cut  by  the  plane  TYS  parallel  to  its  opposite  planes.  There- 
fore as  the  base  ABCD  to  the  base  EFGH,  so  is  the  altitude  MN 
to  the  altitude  MP:  but  MP  is  equal  to  the  altitude  KL<j  where- 
fore as  the  base  ABCD  to  the  base  EFGH,  so  is  the  altitude  MN 
to  the  altitude  KL;  that  is,  the  bases  and  altitudes  of  the  equal 
cylinders  AX,  EO  are  reciprocally  proportional. 

But  let  the  bases  and  altitudes  of  the  cylinders  AX,  EO  be 
reciprocally  proportional,  viz.  the  base  ABCD  to  the  base  EFGH, 
as  the  altitude  MN  to  the  altitude  KL:  the  cylinder  AX  is  equal 
to  the  cylinder  EO. 

First,  let  the  base  ABCD  be  equal  to  the  base  EFGH;  then 
because  as  the  base  ABCD  is  to  the  base  EFGH,  so  is  the  alti- 
tude MN  to  the  altitude  KL;  MN  is  equal  (A.  5.)  to  KL,  and 
therefore  the  cylinder  AX  is  equal  (11.  12.)  to  the  cylinder  EO. 

But  let  the  bases  ABCD,  EFGH  be  unequal,  and  let  ABCD 
be  the  greater;  and  because  as  ABCD  is  to  the  base  EFGH,  so 
is  the  altitude  MN  to  the  altitude  KL;  therefore  MN  is  greater 
(A.  5.)  than  KL.  Then,  the  same  construction  being  made  as 
before,  because  as  the  base  ABCD  to  the  base  EFGH,  so  is  the 
altitude  MN  to  the  altitude  KL;  and  because  the  altitude  KL  is 
equal  to  the  altitude  MP;  therefore  the  base  ABCD  is  (11.  12.) 
to  the  base  EFGH,  as  the  cylinder  AX  to  the  cylinder  ES;  and  as 
the  altitude  MN  to  the  altitude  MP  or  KL,  so  is  the  cylinder 
EO  to  the  cylinder  ES:  therefore  the  cylinder  AX  is  to  the  cy- 
linder ES,  as  the  cylinder  EO  is  to  the  same  ES;  whence  the 
cylinder  AX  is  equal  to  the  cylinder  EO:  and  the  same  reasoning 
holds  in  cones.     Q.  E.  D. 


PROP.  XVL  PROB. 

To  describe  in  the  greater  of  the  two  circles  that  have 
the  same  centre,  a  polygon  of  an  even  number  of  equal 
sides,  that  shall  not  meet  the  lesser  circle. 

Let  ABCD,  EFGH  be  two  given  circles  having  the  same  cen- 
tre K:  it  is  required  to  inscribe  in  the  greater  circle  ABCD  a 
polygon  of  an  even  number  of  equal  sides,  that  shall  not  meet 
the  lesser  circle. 

Through  the  centre  K  draw  the  straight  line  BD,  and  from 
the  point  G,  Avhere  it  meets  the  circumference  of  the  lesser 
circle,  draw  GA  at  right  angles  to  BD,  and  produce  it  to  C; 
therefore  AC  touches  (16.  3.)  the  circle  EFGH:  then,  if  the  cir- 
cumference BAD  be  biscctecl,  and  the  half  of  it  be  again  bisect- 
ed, and  so  on,  there  must  at  length  remain  a  circumference  less 


223 


THE  ELEMENTS  OF  EUCLID. 


BOOK  Xll. 


(Lemma.)  than  AD:  let  this  be  LD; 
and  from  the  point  L  draw  LM  per- 
pendicular to  BD,  and  produce  it  to 
N;  and  join  LD,  DN,  Therefore 
LD  is  equal  to  DN;  and  because  LN 
is  parallel  to  AC,  and  that  AC  touches 
the  circle  EFGH,  therefore  LN  does 
not  meet  the  circle  EFGH:  and  much 
less  shall  the  straight  lines  LD,  DN 
meet  the  circle  EFGH,  so  that  if 
straight  lines  equal  to  LD  be  applied  in  the  circle  ABCD  from 
the  point  L  around  to  N,  there  shall  be  described  in  the  circle  a 
polygon  of  an  even  number  of  equal  sides  not  meeting  the  lesser 
circle.     Which  was  to  be  done. 


LEMMA  H. 

If  two  trapeziums  ABCD,  EFGH  be  inscribed  in  the 
circles,  the  centres  of  which  are  the  points  K,  L;  and  if 
the  sides  AB,  DC  be  parallel,  as  also  EF,  HG:  and  the 
other  four  sides  AD,  BC,  EH,  FG  be  all  equal  to  one 
another;  but  the  side  AB  greater  than  EF,  and  DC 
greater  than  HG;  the  straight  line  KA  from  the  centre 
of  the  circle  in  which  the  greater  sides  are,  is  greater 
than  the  straight  line  LE  drawn  from  the  centre  to  the 
circumference  of  the  other  circle. 

If  it  be  possible,  let  KA  be  not  greater  than  LE;  then  KA 
must  be  either  equal  to  it  or  less.  First,  let  KA  be  equal  to  LE: 
therefore  because  in  two  equal  circles,  AD,  BC  in  the  one, 
are  equal  to  EH,  FG  in  the  other,  the  circumferences  AD,  BC 
are  equal  (28.  S.)  to  the  circumferences  EH,  FG;  but  because 
the  straight  lines  AB,  DC  are  respectively  greater  than  EF., 
GH,  the  circumferences  AB,  DC  are  greater  than  EF,  HG: 
therefore  the  whole  circumference  ABCD  is  greater  than  the 
whole  EFGH;  but  it  is  also  equal  to  it,  which  is  impossible: 
therefore  the  straight  line  KA  is  not  equal  to  LE. 

But  let  KA  be  less  than  LE,  and  make  LM  equal  to  KA,  and 
from  the  centre  L,  and  distance  LM,  describe  the  circle  MNOP, 
meeting  the  straight  lines  LE,  LF,  LG,  LH,  in  M,  N,  O,  P;  and 
join  MN,  NO,  OP,  PM,  which  are  respectively  parallel  (2.  6.)  to 
and  less  than  EF,  FG,  GH,  HE:  then  because  EH  is  greater  than 
MP,  AD  is  greater  than  MP;  and  the  circles  ABCD,  MNOP  are 
equal;  therefore  the  circumference  AD  is  greater  than  MP;  for 
the  same  reason,  the  circumference  BC  is  greater  than  NO;  and 
because  the  straight  line  AB  is  greater  than  EF,  which  is  greater 
than  MN,  much  more  is  AB  greater  than  MN:  therefore  the  cir- 
cumference AB  is  greater  than  MN;  and  for  the  same  reason,  the 
circumference  DC  is  greater  than  PO:  therefore  the  whole  cir- 
cumference ABCD  is  greater  than  the  whole  MNOP;  but  it  is 


BOOK  XII.  THE  ELEMENTS  OF  EUCLID.  229 

likewise  equal  to  it,  which  is  impossible:  therefore  KA  is  not  less 
than  LE^  nor  is  it  equal  to  it:  the  straight  line  KA  must  there- 
fore be  greater  than  LE.     Q.  E  D. 


CoR.  And  if  there  be  an  isosceles  triangle,  the  sides  of  which 
ajre  equal  to  AD,  BC,  but  its  base  less  than  AB  the  gi-eater  of  the 
two  sides  AB,  DC;  the  straight  line  KA  may,  in  the  same  man- 
ner, be  demonstrated  to  be  greater  than  the  straight  line  drawn 
from  the  centre  to  the  circumference  of  the  circle  described  about 
the  triangle. 

PROP.  XVII.  PROB. 

To  describe  in  the  greater  of  two  spheres  which  have 
the  same  centre,  a  sohd  polyhedron,  the  superficies  of 
which  shall  not  meet  the  lesser  sphere.* 

Let  there  be  two  spheres  about  the  same  centre  A;  it  is  required 
to  describe  in  the  greater  a  solid  polyhedron,  the  superficies  of 
which  shall  not  meet  the  lesser  sphere. 

Let  the  spheres  be  cut  by  a  plane  passing  through  the  centre; 
the  common  sections  of  it  with  the  spheres  shall  be  circles:  be- 
cause the  sphere  is  described  by  the  revolution  of  a  semicircle 
about  the  diameter  remaining  unmoveable:  so  that  in  whatever 
position  the  semicircle  be  conceived,  the  common  section  of  the 
plane  in  which  it  is  with  the  superficies  of  the  sphere  is  the  cir- 
cumference of  a  circle;  and  this  is  a  great  circle  of  the  sphere, 
because  the  diameter  of  the  sphere,  which  is  likewise  the  diame- 
ter of  the  circle,  is  greater  (15.  3.)  than  any  straight  line  in  the 
circle  or  sphere:  let  then  the  circle  made  by  the  section  of  the 
plane  with  the  greater  sphere  be  BCDE,  and  with  the  lesser 
sphere  be  FGH;  and  draw  the  two  diameters  BD,  CE  at  right  angles 
to  one  another;  and  in  BCDE,  the  greater  of  the  two  circles,  de- 
scribe (16.  12.)  a  polygon  of  an  even  number  of  equal  sides,  not 
meeting  the  lesser  circle  FGH;  and  let  its  sides,  in  BE,  the  fourth 
part  of  the  circle,  be  BK,  KL,  LM,  ME;  join  KA  and  produce  it 
to  N;  and  from  A  draw  AX  at  right  angles  to  the  plane  of  the 
circle  BCDE,  meeting  the  superficies  of  the  sphere  in  the  point 

*  See  Note. 


230 


THE  ELEMENTS  OF  EUCLID. 


BOOK   2^11. 


X;  and  let  planes  pass  through  AX,  and  each  of  the  straight  Jine's 
BE),  KN,  which,  from  what  has  been  said,  shall  produce  great  cir-  , 
cles  on  the  superficies  of  the  sphere;  and  let  BXD,  KXN  be  the 
semicircles  thus  made  upon  the  diameters  BD,  KN:  therefore, 
because  XA  is  at  right  angles  to  the  plane  of  the  circle  BCDE, 
every  plane  which  passes  through  XA  is  at  right  (18.  1 1.)  angles 
to  the  plane  of  the  circle  BCDE)  wherefore  the  semicircles  BXD, 
KXN  are  at  right  angles  to  that  plane;  and  because  the  semi- 
circles BED,  BXD,  KXN,  upon  the  equal  diameters  BD,  KN  are 
equal  to  one  another,  their  halves  BE,  BX,  KX,  are  equal  to  one 
another;  therefore,  as  many  sides  of  the  polygon  as  are  in  BE,  so 
many  there  are  in  BX,  KX  equal  to  the  sides  BK,  KL,  LM, 
ME:  let  these  polygons  be  described,  and  their  sides  be  BO,  OP, 
PR,  RX;  KS,  ST,  TY,  YX,  and  join  OS,  PT,  RY;  and  from  the 
points  O,  S,  draw  OV,  SQ  perpendiculars  to  AB,  AK:  and  be- 
cause the  plane  BOXD  is  at  right  angles  to  the  plane  BCDE,  and 
in  one  of  them  BOXD,  OV  is  drawn  perpendicular  to  AB  the 
common  section  of  the  planes,  therefore  OV  is  perpendicular 
(4.  def.  11.)  to  the  plane  BCDE:  for  the  same  reason  SQ  is  per- 
pendicular to  the  same  plane,  because  the  plane  KSXN  is  at  right 
angles  to  the  plane  BCDE.  Join  VQ;  and  because  in  the  equal 
semicircles  BXD,  KXN  the  circumferences  BO,  KS  are  equal. 


and  OV,  SQ  are  perpendicular  to  their  diameters,  therefore  (26. 
1.)  OV  is  equal  to  SQ,  and  BV  equal  to  KQ:  but  the  whole  BA 
is  equal  to  the  whole  KA,  therefore  the  remainder  VA  is  equal  to 


riOOK  XII. 


THE  ELEMENTS  OF  EUCLID. 


231 


the  remainder  QA;  as  therefore  BV  is  to  VA,  so  is  KQ  to  QA, 
wherefore  VQ  is  parallel  (2.  6.)  to  BK|  and  because  OV,  SQ  are 
each  of  them  at  right  angles  to  the  plane  of  the  circle  BCDE,  OV 
is  parallel  (6.  11.)  to  SQ;  and  it  has  been  proved  that  it  is  also 
equal  to  it;  therefore  QV,  SO  are  equal  and  parallel  (33.  1.):  and^ 
because  QV  is  parallel  to  SO,  and  also  to  KB,  OS  is  parallel  (9.' 
11.)  to  BK;  and  therefore  BO,  KS  which  join  them  are  in  the 
same  plane  in  which  these  parallels  are,  and  the  quadrilateral 
figure  KBOS  is  in  one  plane;  and  if  PB,  TK  be  joined,  and  per- 
pendiculars be  drawn  from  the  points  P,  T  to  the  straight  lines 
AB,  AK,  it  may  be  demonstrated,  that  TP  is  parallel  to  KB  in  the 
very  same  way  that  SO  was  shown  to  be  parallel  to  the  same  KB; 
wherefore  (9.  11.)  TP  is  parallel  to  SO,  and  the  quadrilateral 
figure  SOPT  is  in  one  plane:  for  the  same  reason,  the  quadrilate- 
ral TPRY  is  in  one  plane;  and  the  figure  YRX  is  also  in  one  plane 
(2.  11.).     Therefore,  if  from  the  points  O,  S,  P,  T,  R,  Y  there 


be  drawn  straight  lines  to  the  point  A,  there  shall  be  formed  a 
solid  polyhedron  between  the  circumferences  BX,  KX  composed 
of  pyramids,  the  bases  of  which  are  the  quadrilaterals  KBOS, 
SOPT,  TPRY,  and  the  triangle  YRX,  and  of  which  the  common 
vertex  is  the  point  A:  and  if  the  same  construction  be  made  upon 
each  of  the  sides  KL,  LM,  ME,  as  has  been  done  upon  BK,  and 
the  like  be  done  also  in  the  other  three  quadrants,  and  in  the 
other  hemisphere;  there  shall  be  formed  a  solid  polyhedron  de- 
scribed in  the  sphere,  composed  of  pyramids,  the  bases  of  which 


232  THE  ELEMENTS  OF  EUCLID.  BOOIC  XII. 

are  the  aforesaid  quadrilateral  figures  and  the  triangle  YRX,  and 
those  formed  in  the  like  manner  in  the  rest  of  the  sphere,-  the 
common  vertex  of  them  all  being  the  point  A;  and  the  superficies 
of  this  solid  polyhedron  does  not  meet  the  lesser  sphere  in  which 
^s  the  circle  FGH:  for,  from  the  point  A  draw  (ll.  11.)  AZ  per- 
pendicular to  the  plane  of  the  quadrilateral  KBOS,  meeting  it  in 
Z,  and  join  BZ,  ZK:  and  because  AZ  is  perpendicular  to  the 
plane  KBOS,  it  makes  right  angles  with  every  straight  line  meet- 
ing it  in  that  planej  therefore  AZ  is  perpendicular  to  BZ  and  ZK; 
and  because  AB  is  equal  to  AK,  and  that  the  squares  of  AZ,  ZB 
are  equal  to  the  square  of  ABj  and  the  squares  of  AZ,  ZK  to  the 
square  of  AK  (47.  1.):  therefore  the  squares  of  AZ,  ZB  are  equal 
to  the  squares  of  AZ,  ZK:  take  from  these  equals  the  square  of 
AZ,  the  r^^aining  square  of  BZ  is  equal  to  the  remaining  square 
of  ZK^  and  therefore  the  straight  line  BZ  is  equal  to  ZK:  in  the 
like  manner  it  may  be  demonstrated,  that  thfe  straight  lines  drawn 
from  the  point  Z  to  the  points  O,  S,  are  equal  to  BZ  or  ZK:  there- 
fore the  circle  described  from  the  centre  Z,  and  distance  ZB,  shall 
pass  through  the  points  K,  O,  S,  and  KBOS  shall  be  a  quadrilate- 
ral figure  in  the  circle:  and  because  KB  is  greater  than  QV,  and 
and  QV  equal  to  SO,  therefore  KB  is  greater  than  SO5  but  KB  is 
equal  to  each  of  the  straight  lines  BO,  KS^  wherefore  each  of  the 
circumferences  cut  off  by  KB,  BO,  KS  is  greater  than  that  cut  off 
by  OS5  and  these  three  circumferences,  together  with  a  fourth 
equal  to  one  of  them,  are  greater  than  the  same  three  together 
with  that  cut  off  by  OS,  that  is,  than  the  whole  circumference  of 
the  circle;  therefore  the  circumference  subtended  by  KB  is  greater 
than  the  fourth  part  of  the  whole  circumference  of  the  circle  KBOS, 
and  consequently  the  angle  BZK  at  the  centre  is  greater  than  a 
right  angle:  and  because  the  angle  BZK  is  obtuse,  the  square  of 
BK  is  greater  (12.  2.)  than  the  squares  of  BZ,  ZK;  that  is,  greater 
than  twice  the  square  of  BZ.  Join  KV,  and  because  in  the  tri- 
angles KBV,  OBV,  KB,  BV  are  equal  to  OB,  BV,  and  that  they 
contain  equal  angles;  the  angle  KVB  is  equal  (4.  1.)  to  the  angle 
OVB:  and  OVB  is  a  right  angle;  therefore  also  KVB  is  a  right 
angle:  and  because  BD  is  less  than  twice  DV,  the  rectangle  con- 
tained by  DB,  BV  is  less  than  twice  the  rectangle  DVB;  that  is 
(8.  6.),  the  square  of  KB  is  less  than  twice  the  square  of  KV:  but 
the  square  of  KB  is  greater  than  twice  the  square  of  BZ;  there- 
fore the  square  of  KV  is  greater  than  the  square  of  BZ:  and  be- 
cause BA  is  equal  to  AK,  and  that  the  squares  of  BZ,  ZA  are 
equal  together  to  the  square  of  BA,  and  the  squares  of  KV,  VA  to 
the  square  of  AK;  therefore  the  squares  of  BZ,  ZA  are  equal  to 
the  squares  of  KV,  VA;  and  of  these  the  square  of  KV  is  greater 
than  the  square  of  BZ;  therefore  the  square  of  VA  is  less  than 
the  square  of  ZA,  and  the  straight  line  AZ  greater  than  VA: 
much  more  then  is  AC  greater  than  AG;  because,  in  the  preced- 
ing proposition,  it  was  shown  that  KV  falls  without  the  circle 
FGH:  and  AZ  is  perpendicular  to  the  plane  KBOS,  and  is  there- 
fore the  shortest  of  all  the  straight  lines  that  can  be  drawn  from 
A,  the  centre  of  the  sphere,  to  that  plane.  Therefore  the  plane 
KBOS  does  not  meet  the  lesser  sphere. 


BOOK  XII.  THE  ELEMENTS  OF  EUCLID,  233 

AticI  that  the  other  planes  between  the  quadrants  BX,  KX  fall 
without  the  lesser  sphere,  is  thus  demonstrated ^  from  the  point  A 
draw  AI  perpendicular  to  the  plane  of  the  quadrilateral  SOFT, 
and  join  lO;  and,  as  was  demonstrated  of  the  plane  KBOS,  and 
the  point  Z,  in  the  same  way  it  may  be  shown  that  the  point  I  is 
the  centre  of  a  circle  described  about  SOPTj  and  that  OS  is 
greater  than  PT;  and  PT  was  shown  to  be  parallel  to  OS;  there- 
fore, because  the  two  trapeziums  KBOS,  SOPT  inscribed  in  cir- 
cles have  their  sides  BK,  OS,  parallel,  as  also  OS,  PT;  and  their 
other  sides  BO,  KS,  OP,  ST  all  equal  to  one  another,  and  that 
BK  is  greater  than  OS,  and  OS  greater  than  PT,  therefore  the 
straight  line  ZB  is  greater  (2.  lem.  12.)  than  lO.  Join  AO  which 
will  be  equal  to  AB:  and  because  AIO,  AZB  are  right  angles,  the 
squares  of  AI,  TO  are  equal  to  the  square  of  AO  or  of  AB;  that 
is,  to  the  squares  of  AZ,  ZB:  and  the  square  of  ZB  is  greater 
than  the  square  of  lO,  therefore,  the  square  of  AZ  is  less  than  the 
square  of  AI;  and  the  straight  line  AZ  less  than  the  straight  line 
AI;  and  it  was  proved  that  AZ  is  greater  than  AG:  much  more 
then  is  AI  greater  than  AG:  therefore  the  plane  SOPT  falls  wholly 
without  the  lesser  sphere:  in  the  same  manner  it  may  be  demon- 
strated that  the  plane  TPRY  falls  without  the  same  sphere,  as  also 
the  triangle  YRX,  viz.  by  the  cor.  of  2d  lemma.  And  after  the 
same  way  it  may  be  demonstrated  that  all  the  planes  which  con- 
tain the  solid  polyhedron,  fall  without  the  lesser  sphere.  There- 
fore in  the  greater  of  two  spheres  which  have  the  same  centre,  a 
solid  polyhedron  is  described,  the  superficies  of  which  does  not 
meet  the  lesser  sphere.     Which  was  to  be  done. 

But  the  straight  line  AZ  may  be  demonstrated  to  be  greater 
than  AG,  otherwise,  and  in  a  shorter  manner,  without  the  help  of 
prop.  16,  as  follows.  From  the  point  G  draw  GU  at  right  angles 
to  AG,  and  join  AU.  If  then  the  circumference  BE  be  bisected, 
and  its  half  again  bisected,  and  so  on,  there  will  at  length  be  left 
a  circumference  less  than  the  circumference  which  is  subtended 
by  a  straight  line  equal  to  GU,  inscribed  in  the  circle  BCDE:  let 
let  this  be  the  circumference  KB;  therefore  the  straight  line  KB 
is  less  than  GU;  and  because  the  angle  BZK  is  obtuse,  as  was 
proved  in  the  preceding,  therefore  BK  is  greater  than  BZ:  but 
GU  is  greater  than  BK;  much  more  then  is  GU  greater  than  BZ, 
and  the  square  of  GU  than  the  square  of  BZ,  and  AU  is  equal  to 
AB;  therefore  the  square  of  AU,  that  is,  the  squares  of  AG,  GU 
are  equal  to  the  square  of  AB,  that  is,  to  the  squares  of  AZ,  ZB; 
but  the  square  of  BZ  is  less  than  the  square  of  GU;  therefore  the 
square  of  AZ  is  greater  than  the  square  of  AG,  and  the  straight 
line  AZ  consequently  greater  than  the  straight  line  AG. 

Cor.  And  if  in  the  lesser  sphere  there  be  described  a  solid  poly- 
hedron, by  drawing  straight  lines  betwixt  the  points  in  which  the 
straight  lines  from  the  centre  of  the  sphere  drawn  to  all  the  angles 
of  the  solid  polyhedron  in  the  greater  sphere  meet  the  superficies 
of  the  lesser;  in  the  same  order  in  which  are  joined  the  points  in 
which  the  same  lines  from  the  centre  meet  the  superficies  of  the 
greater  sphere;  the  solid  polyhedron  in  the  sphere  BCDE  has  to 
this  other  solid  polyhedron  the  tri[)licate  ratio  of  that  which  the 
2  G 


234 


THE  ELEMENTS  OF  EUCLID. 


BOOK  XII. 


diameter  of  the  sphere  BCDE  has  to  the  diameter  of  the  other 
sphere;  for  if  these  two  solids  be  divided  into  the  same  number  of 
pyramids,  and  in  the  same  order,  the  pyramids  shall  be  similar  to 
one  another  each  to  each;  because  they  have  the  solid  angles  at 
their  common  vertex,  the  centre  of  the  sphere  the  same  in  each 
pyramid,  and  their  other  solid  angle  at  the  bases  equal  to  one  ano- 
ther, each  to  each  (B.  1 1.),  because  they  are  contained  by  three  plane 
angles  equal  each  to  each:  and  the  pyramids  are  contained  by  the 
same  number  of  similar  planes;  and  are  therefore  similar  (11.  def. 
11.)  to  one  another,  each  to  each:  but  similar  pyramids  have  to 
one  another  the  triplicate  (Cor.  8.  12.)  ratio  of  their  homologous 
sides.  Therefore  the  pyramid  of  which  the  base  is  the  quadri- 
lateral KBOS,  and  vertex  A,  has  to  the  pyramid  in  the  other 
sphere  of  the  same  order,  the  triplicate  ratio  of  their  homologous 
sides,  that  is  of  that  ralio  which  AB  from  the  centre  of  the  greater 
sphere  has  to  the  straight  line  from  the  same  centre  to  the  super- 
ficies of  the  lesser  sphere.  And  in  like  manner,  each  pyramid  in 
the  greater  sphere  has  to  each  of  the  same  order  in  the  lesser,  the 
triplicate  ratio  of  that  which  AB  has  to  the  semidiameter  of  the 
lesser  sphere.  And  as  one  antecedent  is  to  its  consequent,  so  are 
all  the  antecedents  to  all  the  consequents.  Wherefore  the  whole 
solid  polyhedron  in  the  greater  sphere  has  to  the  whole  solid  poly- 
hedron in  the  other,  the  triplicate  ratio  of  that  which  AB,  the 
semidiameter  of  the  first,  has  to  the  semidiameter  of  the  other; 
that  is,  which  the  diameter  BD  of  the  greater  has  to  the  diameter 
of  the  other  sphere. 

PROP.  XVIII.  THEOR. 

Spheres  have  to  one  another  the  triphcate  ratio  of 
that  which  their  diameters  have. 

Let  ABC,  DEF  be  two  spheres,  of  which  the  diameters  are  BC, 
EF.  The  sphere  ABC  has  to  the  sphere  DEF  the  triplicate  ratio 
of  that  which  BC  has  to  EF. 

For,  if  it  has  not,  the  sphere  ABC  shall  have  to  a  sphere  either 
less  or  greater  than  DEF,  the  triplicate  ratio  of  that  which  BC 
has  to  EF.  First,  let  it  have  that  ratio  to  a  less,  viz.  to  the  sphere 
GHK;  and  let  the  sphere  DEF  have  the  same  centre  with  GHK; 
and  in  the  greater  sphere  DEF  describe  ;(17.  12,)  a  solid  polyhe- 

A  D  1. 


CE 


fm 


dron,  the  superficies  of  which  does  not  meet  the  lesser  sphere 
GHK;  and  in  the  sphere  ABC  describe  another  similar  to  that  in 


BOOK  XII.  THE  ELEMENTS  OF  EUCLID.  235 

the  sphere  DEF;  therefore  the  solid  polyhedron  in  the  sphere 
ABC  has  to  the  solid  polyhedron  in  the  sphere  DEF,  the  tripli- 
cate ratio  (Cor.  17.  12.)  of  that  which  BC  has  to  EF.     But  the 
sphere  ABC  has  to  the  sphere  GHK,  the  triplicate  ratio  of  that 
which  BC  has  to  EF:  therefore,  as  the  sphere  ABC  to  the  sphere 
GHK,  so  is  the  solid  polyhedron  in  the  sphere  ABC  to  the  solid 
polyhedron  in  the  sphere  DEF:  but  the  sphere  ABC  is  greater 
than  the  solid  polyhedron  in  itj  therefore  (14.  5.)  also  the  sphere 
GHK  is  greater  than  the  solid  polyhedron  in  the  sphere  DEF; 
but  it  is  also  less,  because  it  is  contained  within  it,  which  is  im- 
possible: therefore  the  sphere  ABC  has  not  to  any  sphere  less 
than  DEF,  the  triplicate  ratio  of  that  which  BC  has  to  EF.     In 
the  same  manner,  it  may  be  demonstrated  that  the  sphere  DEF 
has  not  to  any  sphere  less  than  ABC,  the  triplicate  ratio  of  that 
which  EF  has  to  BC.     Nor  can  the  sphere  ABC  have  to  any 
sphere  greater  than  DEF,  the  triplicate  ratio  of  that  which  BC 
has  to  EF:  for,  if  it  can,  let  it  have  that  ratio  to  a  greater  sphere 
LMN:  therefore,  by  inversion,  the  sphere  LMN  has  to  the  sphere 
ABC  the  triplicate  ratio  of  that  which  the  diameter  EF  has  to  the 
diameter  BC.     But  as  the  sphere  LMN  to  ABC,  so  is  the  sphere 
DEF  to  some  sphere,  which  must  be  less  (14.  5.)  than  the  sphere 
ABC,  because  the  sphere  LMN  is  greater  than  the  sphere  DEF. 
Therefore  the  sphere  DEF  has  to  a  sphere  less  than  ABC  the  tri- 
,  plicate  ratio  of  that  which  EF  has  to  BC;  which  was  shown  to  be 
impossible:  therefore  the   sphere  ABC  has   not  to   any  sphere 
greater  than  DEF  the  triplicate  ratio  of  that  which  BC  has  to  EF: 
and  it  was  demonstrated,  that  neither  has  it  that  ratio  to  any 
sphere  less  than  DEF.     Therefore  the  sphere  ABC  has  to  the 
sphere  DEF,  the  triplicate  ratio  of  that  which  BC  has  to  EF.    Q. 
E.  D. 


FINIS. 


IVOTES, 


CRITICAL  AND  GEOMETRICAL; 


CONTAINING 


AN  ACCOUNT  OF  THOSE  THINGS  IN  WHICH  THIS  EDITION 
DIFFERS  FROM  THE  GREEK  TEXT : 


AND 

IHE  REASONS  OF  THE  ALTERATIONS  WHICH  HAVE  BEEN  MADE. 

AS  ALSO 

OBSERVATIONS 

ojs  some  of  the  propositions, 

by  robert  simson,  m.  b. 

EMEKIXDS  PnOi'ESSOll  Of  MATHKMATICS  IN  THE  UNIVEIISITY  Of  GJUASGOW- 


.^ 


I 


NOTES, 


^^•*♦^T©  ^5  ©^♦^•^■^ 


DEFINITION  I.     BOOK  I. 

It  is  necessary  to  consider  a  solid,  that  is,  a  magnitude  which 
has  length,  breadth,  and  thickness,  in  order  to  understand  aright 
the  definitions  of  a  point,  line,  and  superficies^  for  these  all  arise 
from  a  solid,  and  exist  in  it:  the  boundary,  or  boundaries  which 
contain  a  solid  are  called  superficies,  or  the  boundary  which  is 
common  to  two  solids  which  are  contiguous,  or  which  divides 
one  solid  into  two  contiguous  parts,  is  called  a  superficies:  thus, 
if  BCGF  be  one  of  the  boundaries  which  contain  the  solid 
ABCDEFGH,  or  which  is  the  common  boundary  of  this  solid, 
and  the  solid  BKLCFNMG,  and  is  therefore  in  the  one  as  well  as 
in  the  other  solid  called  a  superficies,  and  has  no  thickness:  for, 
if  it  have  any,  this  thickness  must  either  be  a  part  of  the  thick- 


H 


G 


M 


F 

/ 

/ 

K       /^ 

Vi 

c 

i 

1 

/ 

< 

/ 

/ 

L 


A 


B 


K 


ness  of  the  solid  AG,  or  the  solid 
BM,  or  a  part  of  the  thickness  of 
each  of  them.  It  cannot  be  a  part 
of  the  thickness  of  the  solid  BM^ 
because,  if  this  solid  be  removed 
from  the  solid  AG,  the  superficies 
BCGF,  the  boundary  of  the  solid 
AG,  remains  still  the  same  as  it 
was.  Nor  can  it  be  a  part  of  the 
thickness  of  the  solid  AGj  because 
if  this  be  removed  from  the  solid  BM,  the  superficies  BCGF, 
the  boundary  of  the  solid  BM,  does  nevertheless  remain:  there- 
fore the  superficies  BCGF  has  no  thickness,  but  only  length  and 
breadth. 

The  boundary  of  a  superficies  is  called  a  line,  or  a  line  is  the 
common  boundary  of  two  superficies  that  are  contiguous,  or 
which  divides  one  superficies  into  two  contiguous  parts:  thus,  if 
BC  be  one  of  the  boundaries  which  contain  the  superficies,  ABCD, 
or  which  is  the  common  boundary  of  this  superficies,  and  of  the 
superficies  KBCL,  which  is  contiguous  to  it,  this  boundary  BC 
is  called  a  line,  and  has  no  breadth:  for  if  it  have  any,  this  must 
be  part  either  of  the  breadth  of  the  superficies  ABCE),  or  of  the 
superficies  KBCL,  or  a  part  of  each  of  them.  It  is  not  part  of 
the  breadth  of  the  superficies  KBCL;  for,  if  these  superficies  be 
removed  from  the  superficies  ABCD,  the  line  BC,  which  is  the 


240 


NOTES. 


BOOK  I. 


boundary  of  the  superficies  ABCD,  remains  the  same  as  it  was: 
nor  can  the  breadth  that  BC  is  supposed  to  have,  be  a  part  of  the 
breadth  of  the  superficies  ABCDj  because,  if  this  be  removed 
from  the  superficies  KBCL,  the  line  BC,  which  is  the  boundary 
of  the  superficies  KBCL,  does  nevertheless  remain:  therefore  the 
line  BC  has  no  breadth:  and  because  the  line  BC  is  a  superficies, 
and  that  a  superficies  has  no  thickness,  as  was  shov/n;  therefore 
a  line  has  neither  breadth  nor  thickness,  but  only  length. 

The  boundary  of  a  line  is  called  a  point,  or  a  point  is  the  com- 
mon boundary  or  extremity  of  two  H  G  M 
lines  that  are  contiguous:  thus,  if 
B  be  the  extremity  of  the  line  AB,  E 
or  the  common  extremity  of  the 
two  lines  AB,  KB,  this  extremity 
is  called  a  point,  and  has  no  length; 
for,  if  it  have  any,  this  length  must 
either  be  part  of  the  length  of  the 
line  AB,  or  of  the  line  KB.  It  is  A  B  K 
not  part  of  the  length  of  KB:  for  if  the  line  KB  be  removed  from 
AB,  the  point  B,  which  is  the  extremity  of  the  line  AB,  remains 
the  same  as  it  was:  nor  is  it  part  of  the  length  of  the  line  AB; 
for,  if  AB  be  removed  from  the  line  KB,  the  point  B,  which  is 
the  extremity  of  the  line  KB,  does  nevertheless  remain:  therefore 
the  point  B  has  no  length:  and  because  a  point  is  in  a  line,  and  a 
line  has  neither  breadth  nor  thickness,  therefore  a  point  has  no 
length,  breadth,  nor  thickness.  And  in  this  manner  the  defini- 
tions of  a  point,  line,  and  superficies,  are  to  be  understood. 


/ 


L,/ 

N 

/ 

< 

■ 

C 

\7 

/ 

DEF.  VII.     B.  I. 

Instead  of  this  definition  as  it  is  in  the  Greek  copies,  a  more 
distinct  one  is  given  from  a  property  of  a  plane  superficies,  which 
is  manifestly  supposed  in  the  Elements,  viz.  that  a  straight  line 
drawn  from  any  point  in  a  plane  to  any  other  in  it  is  wholly  in 
that  plane. 

DEF.  VIII.     B.  I. 

It  seems  that- he  who  made  this  definition  designed  that  it 
should  comprehend  not  only  a  plane  angle  contained  by  two 
straight  lines,  but  likewise  the  angle  which  some  conceive  to  be 
made  by  a  straight  line  and  a  curve,  or  by  two  curve  lines,  which 
meet  one  another  in  a  plane:  but,  though  the  meaning  of  the 
words  £7r'  iv^iiac,^  that  is,  in  a  straight  line,  or  in  the  same  direc- 
tion, be  plain,  when  two  straight  lines  are  said  to  be  in  a  straight 
line,  it  does  not  appear  what  ought  to  be  understood  by  these 
words,  when  a  straight  line  and  a  curve,  or  two  curve  lines,  are 
said  to  be  in  the  same  direction;  at  least  it  cannot  be  explained 
in  this  place;  which  makes  it  probable  that  this  definition,  and 
that  of  the  angle  of  a  segment,  and  what  is  said  of  the  angle  of  a 
semicircle,  and  the  angles  of  segments,  in  the  IGlh  and  31st  pro- 
positions of  book  3,  are  the  additions  of  some  less  skilful  editor: 
on  which  account,  especially  since  they  are  quite  useless,  these 


^^ 


BOOK  I.  NOTES.  241 

definitions  are  dislinguished  from  the  rest  by  inverted  double 
commas. 

DEF.  XVII.  B.  I. 

The  words, "which  also  divides  the  circle  into  two  equal  parts,** 
are  added  at  the  end  of  this  definition  in  all  the  copies,  but  are 
-  now  left  out  as  not  belonging  to  the  definition,  being  only  a  co- 
rollary from  it.  Proclus  demonstrates  it  by  conceiving  one  of 
the  parts  into  which  the  diameter  divides  the  circle,  to  be  applied 
to  the  other;  for  it  is  plain  they  must  coincide,  else  the  straight 
lines  from  the  centre  to  the  circumference  would  not  be  all  equal: 
the  same  thing  is  easily  deduced  from  the  31st  prop,  of  book  3, 
and  the  24th  of  the  same;  from  the  first  of  v/hich  it  follows,  that 
semicircles  ai^e  similar  segments  of  a  circle:  and  from  the  other, 
that  they  are  equal  to  one  another. 

DEF.  XXXIII.  B.  I. 

This  definition  has  one  condition  more  than  is  necessary;  be- 
cause every  quadrilateral  figure  which  has  its  opposite  sides  equal 
to  one  another,  has  likewise  its  opposite  angles  equal,  and  on  the 
contrary. 

Let  ABCD  be  a  quadrilateral  figure,  of  which  the  opposite 
sides  AB,  CD  are  equal  to  one  another:         j^  j) 

as  also  AD  and  BC:  join  BD;  the  two 
sides  AD,  DB  are  equal  to  the  two  CB, 
BD,  and  the  base  AB  is  equal  to  the  base 
CD,  therefore,  by  prop.  8,  of  book  1,  the 
angle  ADB  is  equal  to  the  angle  CBD;     -^  ^ 

and,  by  prop.  4,  book  1,  the  angle  BAD  is  equal  to  the  angle  DCB, 
and  ABD  to  BDC;  and  therefore  also  the  angle  ADC  is  equal  to 
the  angle  ABC. 

And  if  the  angle  BAD  be  ^qual  to  the  opposite  angle  BCD, 
and  the  angle  ABC  to  ADC,  the  opposite  sides  are  equal;  be- 
cause, by  prop.  32,  book  1,  all  the  angles  of  the  quadrilateral 
figure  ABCD  are  together  equal  to  four  A  D 

right  angles,  and  the  two  angles  BAD, 
ADC,  are  together  equal  to  thl  two  angles 
BCD,  ABC:  wherefore  BAD^  ADC  are 
the  half  of  all  the  four  angles;  that  is  BAD 
and  ADC  are  equal  to  two  right  angles:       B  C 

and  therefore  AB,  CD  are  parallels  by  prop.  28,  B.  1.  In  the 
same  manner,  AD,  BC  are  parallels:  therefore  ABCD  is  a  paral- 
lelogram, and  its  opposite  sides  are  equal,  by  prop.  34,  book  h 

PROP.  VII.  B.  I. 

There  are  two  cases  of  this  proposition,  one  of  which  is  not 
in  the  Greek  text,  but  it  is  as  necessary  as  the  other:  and  that 
the  case  left  out  has  been  formerly  in  the  text,  appears  plainly 
from  this,  that  the  second  part  of  prop.  5,  which  is  necessary  to 
the  demonstration  of  this  case,  can  be  of  no  use  at  all  in  the  Ele- 
2  H 


^ 


242  NOTES.  BOOK  I. 

merits,  or  anywhere  else,  but  in  this  demonstration:  because  the 
second  part  of  prop.  5,  clearly  follows  from  the  first  part,  and 
prop.  13,  book  1.  This  part  must  therefore  have  been  added  to 
prop.  5,  upon  account  of  some  proposition  betwixt  the  5th  and 
13th,  but  none  of  these  stand  in  need  of  it  except  the  7th  proposi- 
tion, on  account  of  which  it  has  been  added:  besides,  the  trans- 
lation from  the  Arabic  has  this  case  explicitly  demonstrated. 
And  Proclus  acknowledges,  that  the  second  part  of  prop.  5,  was 
added  upon  account  of  prop.  7,  but  gives  a  ridiculous  reason  for 
it,  "that  it  might  afford  an  answer  to  objections  made  against 
the  7th,"  as  if  the  case  of  the  7th,  which  is  left  out,  were,  as  he 
expressly  makes  it,  an  objection  against  the  proposition  itself. 
Whoever  is  curious,  may  read  what  Proclus  says  of  this  in  his 
commentary  on  the  5th  and  7th  propositions:  for  it  is  not  worth 
while  to  relate  his  trifles  at  full  length. 

It  was  thought  proper  to  change  the  enunciation  of  this  7th 
prop,  so  as  to  preserve  the  very  same  meaning;  the  literal  trans- 
lation from  the  Greek  being  extremely  harsh,  and  difficult  to  be 
understood  by  beginners. 

PROP.  XI.  B.  I. 

A  corollary  is  added  to  this  proposition,  which  is  necessary  to 
prop.  1.  B.  11,  and  otherwise. 

PROP.  XX.  and  XXI.  B.  I. 

Proclus,  in  his  commentary,  relates,  that  the  Epicureans  de- 
rided this  proposition,  as  being  manifest  even  to  asses,  and  needing 
no  demonstration;  and  his  answer  is,  that  though  the  truth  of  it 
be  manifest  to  our  senses,  yet  it  is  science  which  must  give  the 
reason  why  two  sides  of  a  triangle  are  greater  than  the  third:  but 
the  right  answer  to  this  objection  against  this  and  the  21st  and 
some  other  plain  propositions,  is,  that  the  number  of  axioms  ought 
not  to  be  increased  without  necessity,  as  it  must  be  if  these  pro- 
positions be  not  demonstrated.  Mons.  Clairault,  in  the  preface  to 
his  Elements  of  Geometry,  published  in  French  at  Paris  anno 
1741,  says.  That  Euclid  has  been  at  the  pains  to  prove,  that  the 
two  sides  of  a  triangle  which  is  included  within  another  are  to- 
gether less  than  the  two  sides  of  the  triangle  which  includes  it; 
but  he  has  forgot  to  add  this  condition,  viz.  that  the  triangles 
must  be  upon  the  same  base:  because,  unless  this  be  added,  the 
sides  of  the  included  triangle  may  be  greater  than  the  sides  of 
the  triangle  which  includes  it,  in  any  ratio  which  is  less  than  that 
of  two  to  one,  as  Pappus  Alexandrinus  has  demonstrated,  in  prop. 
3.  B.  3,  of  his  mathematical  collections. 

PROP.  XXII.  B.  I. 

Some  authors  blame  Euclid,  because  he  does  not  demonstrate 
that  the  two  circles  made  use  of  in  the  construction  of  this  pro- 
blem must  cut  one  another:  but  this  is  very  plain  from  the  deter- 
mination he  has  given,  viz.  that  any  two  of  the  straight  lines  DF, 


BOOK  I. 


NOTES. 


243 


FG,  GH,  must  be  greater  than  the 
third:  ,for  who  is  so  dull,  though 
only  beginning  to  learn  the  Elements, 
as  not  to  perceive  that  the  circle  de- 
scribed from  the  centre  F,  at  the 
distance  FD,  must  meet  FH  be- 
twixt F  and  H,  because  FD  is  less  DM  F  G  H 
than  FH;  and  that,  for  the  like  reason,  the  circle  described  from 
the  centre  G,  at  the  distance  GH,  or  GM,  must  meet  DG  be- 
twixt D  and  G;  and  that  these  circles  must  meet  one  another,  be- 
cause FD  and  GH  are  together  greater  than  FG?  and  this  deter- 
mination is  easier  to  be  understood 
than  that  which  Mr.  Thomas  Sim- 
son  derives  from  it,  and  puts  in-  ^^  ^ 
stead  of  Euclid's  in  the  49th  page 
of  his  Elements  of  Geometry,  that        •  ,                                v 

he   may   supply   the    omission    he       /_/;__ ]_ 

blames  Euclid  for;  which  determi-     D  M  F  G~  H 

nation  is  that  any  of  the  three  straight  lines  must  be  less  than  the 
sum,  but  greater  than  the  difference  of  the  other  two:  from  this 
he  shows  the  circles  must  meet  one  another,  in  one  case;  and  says 
that  it  may  be  proved  after  the  same  manner  in  any  other  case; 
but  the  straight  line  GM,  which  he  bids  take  from  GF,  may  be 
greater  than  it,  as  in  the  figure  here  annexed:  in  which  case  his 
demonstration  must  be  changed  into  another. 

PROP.  XXIV.  B.  I. 

To  this  is  added,  "  of  the  two  sides  DE,  DF,  let  DE  be  that 
which  is  not  greater  than  the  other;"  that  is,  take  that  side  of 
the  two  DE,  DF,  which  is  not  greater  than  the  other,  in  order  to 
make  with  it  the  angle  EDG  equal  to  BAG,     D 
because,  without  this  restriction,  there  might 
be  three  different  cases  of  the  proposition,  as 
Campanus  and  others  make. 

Mr.  Thomas  Simson,  in  p.  262  of  the  se- 
cond edition  of  his  Elements  of  Geometry, 
printed  anno  1760,  observes,  in  his  notes,  that 
it  ought  to  have  been  shown  that  the  point  F 
falls  below  the  line  EG.  This  probably  Eu- 
clid omitted,  as  it  is  very  easy  to  perceive, 
that  DG  being  equal  to  DF,  the  point  G  is 
in  the  circumference  of  a  circle  described  from 
the  centre  D,  at  the  distance  DF,  and  must  be  in  that  part  of  it 
which  is  above  the  straight  line  EF,  because  DG  falls  above  DF, 
the  angle  EDG  being  greater  than  the  angle  EDF. 

PROP.  XXIX.  B.  I. 

The  proposition  which  is  usually  called  the  5th  postulate,  or 
11th  axiom,  by  some  the  12th,  on  which  this  29th  depends,  has 
given  a  great  deal  to  do,  both  to  ancient  and  modern  geometers: 
it  seems  not  to  be  properly  placed  among  the  axioms,  as  indeed 
it  is  not  self-evident;  but  it  may  be  demonstrated  thus: 


244  NOTES.  BOOK  I. 

DEFINITION  L 

The  distance  of  a  point  from  a  straight  line,  is  the  perpendicu- 
lar drawn  to  it  from  the  point. 

DEF.  2. 

One  straight  line  is  said  to  go  nearer  to,  or  further  from  an^ 
other  straight  line,  when  the  distance  of  the  points  of  the  first 
from  the  other  straight  line  becomes  less  or  greater  than  they 
were;  and  two  straight  lines  are  said  to  keep  the  same  distance 
from  one  another,  when  the  distance  of  the  points  of  one  of  them 
from  the  other  is  always  the  same. 

AXIOM. 

A  straight  line  cannot  first  come  nearer  to  another  straight  line, 
and  then  go  further  from  it,  before  it  A — — -^  __B,_— — —  C 
cuts  it;  and,  in  like  manner,  a  straight 
line  cannot  go  further  from  another 
straight  line,  and  then  come  nearer  to 
it;  nor  can  a  straight  line  keep  the  same 

distance  from  another  straight  line,  and  then  come  nearer  to  it, 
or  go  further  from  it;  for  a  straight  line  keeps  always  the  same 
direction. 

For  example,  the  straight  line  ABC  cannot  first  come  nearer 
to  the  straight  line  DE,  as  from  the  B  ^ 

point   A    to    the   point   B,   and   then,    A -^^^  ^ 

from  the  point  B  to  the  point  C,  go     D E 

further  from  the  same   DE:    and,   in     F -j:;- .^  „ 

like   manner,  the  straight  line  FGH 

cannot  go  further  from  DE,  as  from  F  to  G,  and  then,  from  G 
to  H,  come  nearer  to  the  same  DE:  and  so  in  the  last  case,  as  in 
figure  2.     (See  the  figure  above.) 

PROP.  I. 

If  two  equal  straight  lines,  AC,  BD,  be  each  at  right  angles 
to  the  same  straight  line  AB;  if  the  points  C,  D  be  joined  by  the 
straight  line  CD,  the  straight  line  EF  drawn  from  any  point  E 
in  AB  unto  CD,  at  right  angles  to  AB,  shall  be  equal  to  AC,  or 
BD. 

If  EF  be  not  equal  to  AC,  one  of  them  must  be  greater 
than  the  other;  let  AC  be  the  greater;  then,  because  FE  is 
less  than  CA,  the  straight  line  CFD  is  nearer  to  the  straight  line 
AB  at  the  point  F  than  at  the  point  C, 
that  is,  CF  comes  nearer  to  AB  from 
the  point  C  to  F:  but  because  DB  is 
greater  than  FE,  the  straight  line  CFD 
is  further  from  AB  at  the  point  D  than 
at  F,  that  is  FD  goes  further  from 
AB  from  F  to  D:  therefore  the  straight 
line    CFD    first   comes   nearer    to   the 


BOOK  I.  NOTES. 


245 


straight  line  AB,  and  then  goes  further  from  it,  before  it  cuts  it; 
which  is  impossible.  If  FE  be  said  to  be  greater  than  CA,  or 
DB,  the  straight  line  CFD  first  goes  further  from  the  straight 
line  AB,  and  then  comes  nearer  to  it;  which  is  also  impossible. 
Therefore  FE  is  not  unequal  to  AC,  that  is,  it  is  equal  to  it. 

PROP.  IT. 

If  two  equal  straight  lines  AC,  BD  be  each  at  right  angles  to 
the  same  straight  line  AB;  the  straight  line  CD,  which  joins  their 
extremities,  makes  right  angles  with  AC  and  BD. 

Join  AD,  BC;  and  because,  in  the  triangles  CAB,  DBA,  CA, 
AB  are  equal  to  DB,  BA,  and  the  angle  CAB  equal  to  the  angle 
DBA;  the  base  BC  is  equal  (4.  1.)  to  the  base  AD:  and  in 
the  triangles  ACD,  BDC,  AC,  CD,  are  equal  to  BD,  DC,  and 

the  base  AD  is  equal  to  the  base  BC:         C  F D 

therefore  the  angle  ACD  is  equal  (8. 

1.)  to  the  angle  i3DC:  from  any  point 

E  in  AB  draw  EF  unto  CD,  at  right 

angles  to  AB:   therefore  by   prop.   1. 

EF  is  equal  to  AC,  or  BD;  wherefore,        A  E  B 

as  has  been  just  now  shown,  the  angle  ACF  is  equal  to  the  angle 

EFC:  in  the  same  manner,  the  angle  BDF  is  equal  to  the  angle 

EFD;  but  the  angles  ACD,  BDC  are  equal;  therefore  the  angles 

EFC  and  EFD  are  equal,  and  right  angles  (10.  def  1.);  wherefore 

also  the  angles  ACD,  BDC  are  right  angles. 

Cor.  Hence,  if  two  straight  lines  AB,  CD  be  at  right  angles 
to  the  same  straight  line  AC,  and  if  betwixt  them  a  straight  line 
BD  be  drawn  at  right  angles  to  either  of  them,  as  to  AB;  then 
BD  is  equal  to  AC,  and  BDC  is  a  right  angle. 

If  AC  be  not  equal  to  BD,  take  BG  equal  to  AC,  and  join  CG; 
therefore,  by  this  proposition,  the  angle  ACG  is  a  right  angle; 
but  ACD  is  also  a  right  angle;  wherefore  the  angles  ACD,  ACG; 
are  equal  to  one  another,  which  is  impossible.  Therefore  BD  is 
equal  to  AC;  and  by  this  proposition  BDC  is  a  right  angle. 

PROP.  III. 

If  two  straight  lines  which  contain  an  angle  be  produced,  there 
may  be  found  in  either  of  them  a  point  from  which  the  perpen- 
dicular drawn  to  the  other  shall  be  greater  than  any  given  straight 
line. 

Let  AB,  AC  be  two  straight  lines  which  make  an  angle  with  one 
another,  and  let  AD  be  the  given  straight  line;  a  point  may  be 
found  either  in  AB  or  AC,  as  in  AC,  from  which  the  perpendicu- 
lar drawn  to  the  other  AB  shall  be  greater  than  AD. 

In  AC  take  any  point  E,  and  draw  EF  perpendicular  to  AB; 
produce  AE  to  G,  so  that  EG  be  equal  to  AE,  and  produce  FE  to 
H,  and  make  EH  equal  to  FE,  and  join  JIG.  Because,  in  the 
triangles  AEF,  GEH,  AE,  EF  are  equal  to  GE,  EH,  each  to  each, 
and  contain  equal  (15.  1.)  angles,  the  angle  GHE  is  therefore  equal 
(4.  1.)  to  the  angle  AFE,  which  is  a  right  angle:  draw  GK  perpen- 


246 


NOTES. 


BOOK  I. 


O 

D 
P 


A 

F            K             B             IV1 

-- 

E 
H 

-.^ 

^^^"^^ 

C 

dicular  to  AB;  and  because  the  straight  lines  FK,  HG  are  at  right 

angles  to  FH,  and  KG 

at  right  angles  to  FK; 

KG  is  equal  to  FH,  by 

cor.  pr.  2.  that  is,  to  the 

double  of  FE.     In  the 

same  manner,  if  AG  be 

produced  to  L,  so  that 

GL  be  equal  to  AG,  and 

LM  be  drawn  perpendicular  to  AB,  then  LM  is  double  of  GK, 

and  so  on.     In  AD  take  AN  equal  to  FE,  and  AO  equal  to  KG, 

that  is,  to  the  double' of  FE,  or  AN;  also  take  AP  equal  to  LM, 

that  is  to  the  double  of  KG,  or  AO;  and,  let  this  be  done  till  the 

straight  line  taken  be  greater  than  AD;  let  this  straight  line  so 

taken  be  AP ,  and  because  AP  is  equal  to  LM,  therefore  LM  is 

greater  than  AD.    Which  was  to  be  done. 


PROP.  IV, 

If  two  straight  lines  AB,  CD  make  equal  angles  EAB,  ECD 
with  another  straight  line  EAC  towards  the  same  parts  of  it;  AB 
and  CD  are  at  right  angles  to  some  straight  line. 

Bisect  AC  in  F,  and  draw  FG  perpendicular  to  AB:  take  CH 
in  the  straight  line  CD  equal  to  AG,  and  on  the  contrary  side  of 
AC  to  that  on  which  AG  is,  and  join  FH:  therefore  in  the  trian- 
gles AFG,  CFH,  the  sides  FA,  AG  are  equal  to  FC,  CH,  each  to 
each,  and  the  angle  FAG,  that  (15.  1.) 
is  EAB  is  equal  to  the  angle  FCH; 
wherefore  (4.  1.)  the  angle  AGF  is 
equal  to  CHF,  and  AFG  to  the  angle 
CFH:  to  these  last  add  the  common 
angles  AFH;  therefore  the  two  angles 
AFG,  AFH  are  equal  to  the  two  an- 
gles CFH,  HFA,  which  two  last  are 
equal  together  to  two  right  angles  (13. 
1 .),  therefore  also  AFG,  AFH  are  equal 
to  two  right  angles,  and  consequently 
(14.  1.)  GF  and  FH  are  in  one  straight  line.  And  because  AGF 
is  a  right  angle,  CHF  which  is  equal  to  it  is  also  a  right  angle; 
therefore  the  straight  lines  AB,  CD  are  at  right  angles  to  GH, 


PROP.  V. 

If  two  straight  lines  AB,  CD,  be  cut  by  a  third  ACE  so  as  to 
make  the  interior  angles  BAC,  ACD,  on  the  same  side  of  it,  to- 
gether less  than  two  right  angles;  AB  and  CD  being  produced 
shall  meet  one  another  towards  the  parts  on  which  are  the  two 
angles  which  are  less  than  two  right  angles. 

At  the  point  C  in  the  straight  line  CE  make  (23.  1.)  the  angle 
ECF  equal  to  the  angle  EAB,  and  draw  to  AB  the  straight  line 
CG  at  right  angles  to  CF:  then,  because  the  angles  ECF,  EAB 


BOOK  I. 


NOTES.  247 


M  C 

/      -^ 

K 

/ 

"^^^^ 

^z 

/ 

^^-^n 

A 

^^\ 

-^ 

are  ^qual  to  one  another,  and 
that  the  angles  ECF,  FCA 
are  together  equal  (13.  1.)  to 
two  right  angles,  the  angles 
EAB,  FCA  are  equal  to  two 
right  angles.  But  by  the  hy- 
pothesis, the  angles  EAB, 
ACD  are  together  less  than 
two  right   angles;    therefore    ^  Q 

the  angle  FCA  is  greater  than  H 

ACD  and  CD  falls  between 

CF  and  AB:  and  because  CF  and  CD  make  an  angle  with  one 
another,  by  prop.  3,  a  point  may  be  found  in  either  of  them  CD, 
from  which  the  perpendicular  drawn  to  CF  shall  be  greater  than 
the  straight  line  CG.  Let  this  point  be  H,  and  draw  HK  perpen- 
dicular to  CF,  meeting  AB  in  L:  and  because  AB,  CF  contain 
equal  angles  with  AC  on  the  same  side  of  it,  by  prop.  4,  AB  and 
CF  are  at  right  angles  to  the  straight  line  MNO,  which  bisects 
AC  in  N  and  is  perpendicular  to  CF;  therefore,  by  cor.  prop.  2, 
CG  and  KL  which  are  at  right  angles  to  CF  are  equal  to  one 
another;  and  HK  is  greater  than  CG,  and  therefore  is  greater 
than  KL,  and  consequently  the  point  H  is  in  KL  produced. 
Wherefore  the  straight  line  CDH  drawn  betwixt  the  points  C,  H, 
which  are  on  contrary  sides  of  AL,  must  necessarily  cut  the 
straight  line  AL. 

PROP.  XXXV.  B.  L 

The  demonstration  of  this  proposition  is  changed,  because,  if 
the  method  which  is  used  in  it  was  followed,  there  would  be  three 
cases  to  be  separately  demonstrated,  as  is  done  in  the  translation 
from  the  Arabic;  for,  in  the  Elements,  no  case  of  a  proposition 
that  requires  a  different  demonstration,  ought  to  be  omitted.  On 
this  account  we  have  chosen  the  method  which  Mons.  Clairault 
has  given,  the  first  of  any,  as  far  as  I  know,  in  his  Elements,  page 
21,  and  which  afterwards  Mr.  Simson  gives  in  his  page  32.  But 
whereas  Mr.  Simson  makes  use  of  prop.  26,  b.  1,  from  which  the 
equality  of  the  two  triangles  does  not  immediately  follow,  because, 
to  prove  that,  the  4th  of  b.  1,  must  likewise  be  made  use  of,  as  may 
be  seen  in  the  very  same  case  in  the  34th  prop.  b.  1,  it  was  thought 
better  to  make  use  only  of  the  4th  of  b.  1. 

PROP.  XLV.  B.  I. 

The  straight  line  KM  is  proved  to  be  parallel  to  FL  from  the 
33d  prop.:  whereas  KH  is  parallel  to  FG  by  construction,  and 
KHM,  FGL,  have  been  demonstrated  to  be  straight  lines.  A  co- 
rollary is  added  from  Commandine,  as  being  often  used. 

PROP.  XIIL  B.  H. 

In  this  proposition  only  acute  angled  triangles  are  mentioned, 
whereas  it  holds  true  of  every  triangle;  and  the  demonstrations  of 
the  cases  omitted  are  added:  Commandine  and  Clavius  have  like- 
wise given  their  demonstrations  of  these  cases. 


248  NOTES.  BOOK  I. 

PROP.  XIV.     B.  II. 

In  the  demonstration  of  this,  some  Greek  editor  has  ignorantly 
inserted  the  words  "  but  if  not,  one  of  the  two  BE,  ED  is  the 
greater;  let  BE  be  the  greater,  and  produce  it  to  F,"  as  if  it  was 
of  any  consequence  whether  the  greater  or  lesser  be  produced: 
therefore,  instead  of  these  words,  there  ought  to  be  read  only, 
"  but  if  not,  produce  BE  to  F." 

PROP.  I.     B.  III. 

Several  authors,  especially  among  the  modern  mathematicians 
and  logicians,  inveigh  too  severely  against  indirect  or  apagogic 
demonstrations,  and  sometimes  ignorantly  enough,  not  being 
aware  that  there  are  some  things  that  cannot  be  demonstrated 
any  other  way:  of  this  the  present  proposition  is  a  very  clear  in- 
stance, as  no  direct  demonstration  can  be  given  of  it:  because^ 
besides  the  definition  of  a  circle,  there  is  no  principle  or  property 
relating  to  a  circle  antecedent  to  this  problem,  from  which  either 
a  direct  or  indirect  demonstration  can  be  deduced:  wherefore  it 
is  necessary  that  the  point  found  by  the  construction  of  the  pro- 
blem be  proved  to  be  the  centre  of  the  circle,  by  the  help  of  this 
definition,  and  some  of  the  preceding  propositions:  and  because, 
in  the  demonstration,  this  proposition  must  be  brought  in,  viz. 
straight  lines  from  the  centre  of  a  circle  to  the  circumference  are 
equal,  and  that  the  point  found  by  the  construction  cannot  be  as- 
sumed as  the  centre,  for  this  is  the  thing  to  be  demonstrated;  it 
is  manifest  some  other  point  must  be  assumed  as  the  centre;  and 
if  from  this  assumption  an  absurdity  follows,  as  Euclid  demon- 
strates there  must,  then  it  is  not  true  that  the  point  assumed  is 
the  centre;  and  as  any  point  whatever  was  assumed,  it  follows 
that  no  point,  except  that  found  by  the  construction,  can  be  the 
centre,  from  which  the  necessity  of  an  indirect  demonstration  in 
this  case  is  evident. 

PROP.  XIII.     B.  III. 

As  it  is  much  easier  to  imagine  that  two  circles  may  touch  one 
another  within,  in  more  points  than  one,  upon  the  same  side,  than 
upon  opposite  sides;  the  figure  of  that  case  ought  not  to  have 
been  omitted;  but  the  construction  in  the  Greek  text  would  not 
have  suited  with  this  figure  so  well,  because  the  centres  of  the 
circles  must  have  been  placed  near  to  the  circumferences;  on 
which  account  another  construction  and  demonstration  is  given, 
which  is  the  same  with  the  second  part  of  that  which  Campanus 
has  translated  from  the  Arabic,  where,  without  any  reason,  the 
demonstration  is  divided  into  two  parts. 

PROP.  XV.     B.  III. 

The  converse  of  the  second  part  of  this  proposition  is  wanting, 
though  in  the  preceding,  the  converse  is  added,  in  a  like  case, 
both  in  the  enunciation  and  demonstration;  and  it  is  now  added 
in  this.     Besides,  in  the  demonstration  of  the  first  n-^***  of  this 


BOOK  III.  NOTES.  249 

15th,  the  diameter  AD  (see  Commandine's  figure)  is  proved  to  be 
greater  than  the  straight  line  BC,  by  means  of  another  straight 
line  MN;  whereas  it  may  be  better  done  without  it:  on  which 
accounts  we  have  given  a  different  demonstration,  like  to  that 
which  Euclid  gives  in  the  preceding  14th,  and  to  that  which 
Theodosius  gives  in  prop.  6,  b.  1,  of  his  Spherics  in  this  very 
affair. 

PROP.  XVI.     B.  III. 

In  this  we  have  not  followed  the  Greek  nor  the  Latin  transla- 
tion literally,  but  have  given  what  is  plainly  the  meaning  of  this 
proposition,  without  mentioning  the  angle  of  the  semicircle,  or 
that  which  some  call  the  cornicular  angle,  which  they  conceive 
to  be  made  by  the  circumference  and  the  straight  line  which  is  at 
right  angles  to  the  diameter,  at  its  extremity;  which  angles  have 
furnished  matter  of  great  debate  between  some  of  the  modern 
geometers,  and  given  occasion  of  deducing  strange  consequences 
from  them,  which  are  quite  avoided  by  the  manner  in  which  we 
have  expressed  the  proposition.  And  in  like  manner,  we  have 
given  the  true  meaning  of  prop.  31,  b.  3,  without  mentioning  the 
angles  of  the  greater  or  lesser  segments:  these  passages  Vieta, 
with  good  reason,  suspects  to  be  adulterated,  in  the  386th  page 
of  his  Oper.  Math. 

PROP.  XX.     B.  III. 

The  first  words  of  the  second  part  of  this  demonstration, 
"»exA(55(r^ft;  S^ij  ttcc^^iv^*'  are  wrong  translated  by  Mr.  Briggs  and  Dr. 
Gregory  "Rursus  inclinetur;"  for  the  translation  ought  to  be 
"Rursus  infiectatur;"  as  Commandine  has  it:  a  straight  line  is 
said  to  be  inflected  either  to  a  straight  or  curve  line,  when  a 
straight  line  is  drawn  to  this  line  from  a  point,  and  frona  the 
point  in  which  it  meets  it,  a  straight  line  making  an  angle  with 
the  former  is  drawn  to  another  point,  as  is  evident  from  the  90th 
prop,  of  Euclid's  Data:  for  this  the  whole  line  betwixt  the  first 
and  last  points,  is  inflected  or  broken  at  the  point  of  inflection, 
where  the  two  straight  lines  meet.  And  in  the  like  sense  two 
straight  lines  are  said  to  be  inflected  from  two  points  to  a  third 
point,  when  they  make  an  angle  at  this  point;  as  may  be  seen  in 
the  description  given  by  Pappus  Alexandrinus  of  Apollonius's 
books  de  Locis  planis,  in  the  preface  to  his  7th  book:  we  have 
made  the  expression  fuller  from  the  90th  prop,  of  the  Data. 

PROP.  XXI.     B.  III. 

There  are  two  cases  of  this  proposition,  the  second  of  which, 
viz.  when  the  angles  are  in  a  segment  not  greater  than  a  semicir- 
cle, is  wanting  in  the  Greek:  and  of  this  a  more  simple  demon- 
stration is  given  than  that  which  is  in  Commandine,  as  being  de- 
rived only  from  the  first  case,  without  the  help  of  triangles. 

PROP.  XXIII.  and  XXIV.     B.  III. 

In  proposition  24  it  is  demonstrated,  that  the  segment  AEB 
must  coincide  with  the  segment  CFD,  (see  Commandine's  figure) 
and  that  it  cannot  fall  otherwise,  as  CGD,  so  as  to  cut  the  other 

21 


250  NOTES.  BOOK  III. 

cii'cle  in  a  third  point  G,  from  this,  that,  if  it  did,  a  circle  could 
cut  another  in  more  points  than  two:  but  this  ought  to  have  been 
proved  to  be  impossible  in  the  23d  prop,  as  well  as,  that  one  of 
the  segments  cannot  fall  within  the  other:  this  part  then  is  left 
out  in  the  24th,  and  put  in  its  proper  place,  the  23d  proposition. 

PROP.  XXV.     B.  III. 

This  proposition  is  divided  into  three  cases,  of  which  two  have 
the  same  construction  and  demonstration^  therefore  it  is  now  di- 
vided only  into  two  cases. 

PROP.  XXXIII.     B.  III. 

This  also  in  the  Greek  is  divided  into  three  cases,  of  which 
two,  viz.  one  in  which  the  given  angle  is  acute,  and  the  other  in 
which  it  is  obtuse,  have  exactly  the  same  construction  and  de- 
monstration; on  which  account,  the  demonstration  of  the  last 
case  is  left  out  as  quite  superfluous,  and  the  addition  of  some  un- 
skilful editor;  besides,  the  demonstration  of  the  case  when  the 
angle  given  is  a  right  angle,  is  done  a  round  about  way,  and  is 
therefore  changed  to  a  more  simple  one,  as  was  done  by  Clavius. 

PROP.  XXXV.     B.  III. 

As  the  25th  and  33d  propositions  are  divided  into  more  cases, 
so  this  thirty-fiflh  is  divided  into  fewer  cases  than  are  necessary. 
Nor  can  it  be  supposed  that  Euclid  omitted  them  because  they 
are  easy;  as  he  has  given  the  case,  which  by  far  is  the  easiest  of 
them  all,  viz.  that  in  which  both  the  straight  lines  pass  through 
the  centre;  and  in  the  following  proposition  he  separately  de- 
monstrates the  case  in  which  the  straight  line  passes  through  the 
centre,  and  that  in  which  it  does  not  pass  through  the  centre:  so 
that  it  seems  Theon,  or  same  other,  has  thought  them  too  long 
to  insert:  but  cases  that  require  different  demonstrations,  should 
not  be  left  out  in  the  Elements,  as  was  before  taken  notice  of: 
these  cases  are  in  the  translation  from  the  Arabic,  and  are  now 
put  into  the  text. 

PROP.  XXXVII.     B.  III. 

At  the  end  of  this  the  words  "in  the  same  manner  it  may  be 
demonstrated,  if  the  centre  be  in  AC,"  are  left  out  as  the  addition 
of  some  ignorant  editor. 

DEFINITIONS  OF  BOOK  IV. 

When  a  point  is  in  a  straight  line,  or  any  other  line,  this  point 
is  by  the  Greek  geometers  said  otsrTso-^cs/,  to  be  upon,  or  in  that 
line,  and  when  a  straight  line  or  circle  meets  a  circle  any  way, 
the  one  is  said  uTrrec-^ui  to  meet  the  other:  but  when  a  straight 
line  or  circle  meets  a  circle  so  as  not  to  cut  it,  it  is  said  e^«7rre<rS-<«/, 
to  touch  the  circle:  and  these  two  terms  are  never  promiscuously 
used  by  them:  therefore  in  the  fifth  definition  of  book  4,  the  com- 
pound e(pcc7rri}Tcci  must  be  read,  instead  of  the  simple  uTrTfjrecr,  and 


BOOK  IV.  NOTES.  251 

in  the  Ist,  2d,  5d,  and  6th  definitions  in  Commandine's  transla- 
tion, "tangit,"  must  be  read  instead  of  "Contingit;"  and  in  the  2d 
and  3d  definitions  of  book  3,  the  same  change  must  be  made:  but 
in  the  Greek  text  of  propositions  11th,  I2th,  13th,  18th,  19th, 
book  3,  the  compound  verb  is  to  be  put  for  the  simple. 

PROP.  IV.     B.  IV. 

In  this,  as  also  in  the  8th  and  13th  propositions  of  this  book,  it 
is  demonstrated  indirectly,  that  the  circle  touches  a  straight  line: 
whereas  in  the  17th,  33d,  and  37th  propositions  of  book  3,  the 
same  thing  is  directly  demonstrated:  and  this  way  we  have  chosen 
to  use  in  the  propositions  of  this  book,  as  it  is  shorter. 

PROP.  V.     B.  IV. 

The  demonstration  of  this  has  been  spoiled  by  some  unskilful 
hand:  for  he  does  not  demonstrate,  as  is  necessary,  that  the  two 
straight  lines  which  bisect  the  sides  of  the  triangle  at  right  angles 
must  meet  one  another^  and,  without  any  reason,  he  divides  the 
proposition  into  three  cases;  whereas,  one  and  the  same  construc- 
tion and  demonstration  serves  for  them  all,  as  Campanus  has  ob- 
served; which  useless  repetitions  are  now  left  out:  the  Greek 
text  also  in  the  corollary  is  manifestly  vitiated,  where  mention  is 
made  of  a  given  angle,  though  there  neither  is,  nor  can  be  any 
thing  in  the  proposition  relating  to  a  given  angle. 

PROP.  XV.  and  XVI.     B.  IV. 

In  the  corollary  of  the  first  of  these,  the  words  equilateral  and 
equiangular  are  wanting  in  the  Greek:  and  in  prop.  16,  instead 
of  the  circle  ABCD,  ought  to  be  read  the  circumference  ABCD: 
where  mention  is  made  of  its  containing  fifteen  equal  parts. 

DEF.  III.     B.  V. 

Many  of  the  modern  mathematicians  reject  this  definition:  the 
very  learned  Dr.  Barrow  has  explained  it  at  large  at  the  end  of 
his  third  lecture  of  the  year  1666;  in  which  also  he  answers  the 
objections  made  against  it  as  well  as  the  subject  would  allow:  and 
at  the  end  gives  his  opinion  upon  the  whole  as  follows: 

"I  shall  only  add,  that  the  author  had,  perhaps, no  other  design 
in  making  this  definition,  than  (that  he  might  more  fully  explain 
and  embellish  his  subject)  to  give  a  general  and  summary  idea  of 
ratio  to  beginners,  by  premising  this  metaphysical  definition,  to 
the  more  accurate  definitions  of  ratios  that  are  the  same  to  one 
another,  or  one  of  v/hich  is  greater,  or  less  than  tlie  other:  I  call 
it  a  metaphysical,  for  it  is  not  properly  a  mathematical  definition, 
since  nothing  in  mathematics  depends  on  it,  or  is  deduced,  nor, 
as  I  judge,  can  be  deduced  from  it;  and  the  definition  of  analogy, 
which  follows,  viz.  Analogy  is  the  similitude  of  ratios,  is  of  the 
same  kind,  and  can  serve  for  no  purpose  in  mathematics,  but  only 
to  give  beginners  some  general,  though  gross  and  confused  notions 
of  analogy:  but  the  whole  of  the  doctrine  of  ratios,  and  the  whole 
of  mathematics,  depend  upon  the  accurate  mathematical  defini- 


252  NOTES.  BOOK  V. 

tions  which  follow  this:  to  these  we  ought  principally  to  attend, 
as  the  doctrine  of  ratios  is  more  perfectly  explained  by  them: 
this  third  and  others  like  it,  may  be  entirely  spared  without  any 
loss  to  geometry:   as  we  see  in  the  7th  book  of  the  Elements, 
where  the  proportion  of  numbers  to  one  another  is  defined,  and 
treated  of,  yet  without  giving  any  definition  of  the  ratio  of  num- 
bersj  though  such  a  definition  was  as  necessary  and  useful  to  be 
given  in  that  book,  as  in  this:  but  indeed  there  is  scarce  any  need 
of  it  in  either  of  them:  though  I  think  that  a  thing  of  so  general 
and  abstracted  a  nature,  and  thereby  the  more  difficult  to  be  con- 
ceived and  explained,  cannot  be  more  commodiously  defined  than 
as  the  author  has  done;  upon  which  account  I  thought  fit  to  ex- 
plain it  at  large,  and  defend  it  against  the  captious  objections  of 
those  who  attack  it."     To  this  citation  from  Dr.  Barrow  I  have 
nothing  to  add,  except  that  I  fully  believe  the  3d  and  8th  defini- 
tions are  not  Euclid's,  but  added  by  some  unskilful  editor. 

DEF.  XI.  B.  V. 

It  was  necessary  to  add  the  word  "continual"  before  "propor- 
tionals" in  this  definition;  and  thus  it  is  cited  in  the  33d  prop,  of 
book  1 1. 

After  this  definition  ought  to  have  followed  the  definition  of 
compound  ratio,  as  this  was  the  proper  place  for  it;  duplicate  and 
triplicate  ratio  being  species  of  compound  ratio.  But  Theon  has 
made  it  the  5th  def.  of  book  6,  where  he  gives  an  absurd  and  en- 
tirely useless  definition  of  compound  ratio:  for  this  reason  we  have 
placed  another  definition  of  it  betwixt  the  11th  and  12th  of  this 
book,  which,  no  doubt,  Euclid  gave;  for  he  cites  it  expressly  in 
prop.  23,  book  6,  and  which  Clavius,  Herigon,  and  Barrow  have 
likewise  given,  but  they  retain  also  Theon's,  which  they  ought  to 
have  left  out  of  the  Elements. 

DEF.  XIII.  B.  V. 

This,  and  the  rest  of  the  definitions  following,  contain  the  ex- 
plication of  some  terms  which  are  used  in  the  5th  and  following 
books;  which,  except  a  few,  are  easily  enough  understood  from 
the  propositions  of  this  book  where  they  are  first  mentioned:  they 
seem  to  have  been  added  by  Theon,  or  some  other.  However  it 
be,  they  are  explained  something  more  distinctly  for  the  sake  of 
learners. 

PROP.  IV.  B.  V. 

In  the  construction  preceding  the  demonstration  of  this,  the 
words  fit  e'^i'Xh  ^^Y  whatever,  are  twice  wanting  in  the  Greek,  as 
also  in  the  Latin  translations;  and  are  now  added,  as  being  wholly 
necessary. 

Ibid,  in  the  demonstration;  in  the  Greek,  and  in  the  Latin  trans- 
lation of  Commandine,  and  in  that  of  Mr.  Henry  Briggs,  which  was 
published  at  London  in  1620,  together  with  the  Greek  text  of  the 
first  six  books,  which  translation  in  this  place  is  followed  by  Dr. 
Gregory  in  his  edition  of  Euclid,  there  is  this  sentence  following, 
viz.  "  and  of  A  and  C  have  been  taken  equimultiples  K,  L;  and  of 


BOOK  V.  NOTES.  253 


nd  Dj  any  equimultiples  whatever  (ot  erv^^e)  M,  N;"  which  is 
true,  the  words  "  any  whatever,"  ought  to  be  left  out:  and  it 


B  and 

not  true,  the  words  "  any  whatever,"  ought  to  be  lelt  out:  and  it 
is  strange  that  neither  Mr.  Briggs,  who  did  right  to  leave  out 
these  words  in  one  place  of  prop.  1 3,  of  this  book,  nor  Dr.  Gregory, 
who  changed  them  into  the  word  "  some,"  in  three  places,  and 
left  them  out  in  a  fourth  of  that  same  prop.  13,  did  not  also  leave 
them  out  in  this  place  of  prop.  4,  and  in  the  second  of  the  two 
places  where  they  occur  in  prop.  17,  of  this  book,  in  neither  of 
which  they  can  stand  consistent  with  truth:  and  in  none  of  all 
these  places,  even  in  those  which  they  corrected  in  their  Latin 
translation,  have  they  cancelled  the  words  cc  ervyj  in  the  Greek 
text,  as  they  ought  to  have  done. 

The  same  words  ot  erv^,^  are  found  in  four  places  of  prop.  1 1,  of 
this  book,  in  the  first  and  last  of  which  they  are  necessary,  but  in 
the  second  and  third,  though  they  are  true,  they  are  quite  super- 
fluousj  as  they  likewise  are  in  the  second  of  the  two  places  in 
which  they  are  found  in  the  12th  prop,  and  in  the  like  places  of 
prop.  22,  23  of  this  book;  but  are  wanting  in  the  last  place  of 
prop.  23,  as  also  in  prop.  25,  book  1 1. 

COR.  IV.  PROP.  B.  V. 

This  corollary  has  been  unskilfully  annexed  to  this  proposition, 
and  has  been  made  instead  of  the  legitimate  demonstration,  which, 
without  doubt,  Theon,  or  some  other  editor,  has  taken  away,  not 
from  this,  but  from  its  proper  place  in  this  book:  the  author  of  it 
designed  to  demonstrate,  that  if  four  magnitudes  E,  G,  F,  H  be 
proportionals,  they  are  also  proportionals  inversely,  that  is,  G  is 
to  E,  as  H  to  Fj  which  is  true;  but  the  demonstration  of  it  does 
not  in  the  least  depend  upon  this  4th  prop,  or  its  demonstration: 
for,  when  he  says,  "  because  it  is  demonstrated  that  if  K  be  greater 
than  M,  L  is  greater  than  N,"  Sec. — This  indeed  is  shown  in  the 
demonstration  of  the  4th  prop,  but  not  from  this,  that  E,  G,  F,  H 
are  proportionals:  for  this  last  is  the  conclusion  of  the  proposi- 
tion. Wherefore  these  words,  "  because  it  is  demonstrated,"  Sec. 
are  wholly  foreign  to  his  design;  and  he  should  have  proved,  that 
if  K  be  greater  than  M,  L  is  greater  than  N,  from  this,  that  E,  G, 
F,  H  are  proportionals,  and  from  the  5th  def.  of  this  book,  which 
he  has  not;  but  is  done  in  proposition  B,  which  we  have  given  in 
its  proper  place  instead  of  this  corollary;  and  another  corollary  is 
placed  after  the  4th  prop,  which  is  often  of  use;  and  is  necessary 
to  the  demonstration  of  prop.  18  of  this  book. 

PROP.  V.   B.  V. 

In  the  construction  which  precedes  the  demonstration  of  this 
proposition,  it  is  required  that  EB  may  be  the  same  multiple  of 
CG,  that  AE  is  of  CF;  that  is,  that  EB  be  divided  into  as  many 
equal  parts,  as  there  are  parts  in  AE  equal  to  CF:  from  which 
it  is  evident,  that  this  construction  is  not  Euclid's;  for  he  does 
not  show  the  way  of  dividing  straight  lines,  and  far  less  other 
magnitudes,  into  any  number  of  equal  parts,  until  the  9th  propo- 
sition of  book  6;  and  he  never  requires  any  thing  to  be  done  in 


s. 


254  NOTES. 


BOOK  y» 


the  construction,  of  -which  he  had  not  before  given 
the  method  of  doing.     For  this  reason,  we  have     A 
changed  the  construction  to  one,  which,  without      p  G 

doubt,  is  Euclid's,  in  which  nothing  is  required  but 
to  add  a  magnitude  to  itself  a  certain  number  of  q 

times^  and  this  is  to  be  found  in  the  translation  "T 

from  the  Arabic,  though  the  enunciation  of  the  p- 

proposition  and  the  demonstration  are  there  very      p. 
much  spoiled.     Jacobus  Eeletarius,.who  was  the  jq 

first,  as  far  as  I  know,  who  took  notice  of  this  er- 
ror, gives  also  the  right  construction  in  his  edition  of  Euclid,  after 
he  had  given  the  other  which  he  blames.  He  says,  he  would  not 
leave  it  out,  because  it  was  fine,  and  might  sharpen  one's  genius 
to  invent  others  like  it^  whereas,  there  is  not  the  least  difference 
between  the  two  demonstrations,  except  a  single  word  in  the  con- 
struction, which  very  probably  has  been  owing  to  an  unskilftil  li- 
brarian. Clavius  likewise  gives  both  the  ways;  but  neither  he 
nor  Peletarius,  takes  notice  of  the  reason  why  the  one  is  preferable 
to  the  other. 

PROP.  VI.  B.  V. 

There  are  two  cases  of  this  proposition,  of  which  only  the  first 
and  simplest  is  demonstrated  in  the  Greek:  and  it  is  probable 
Theon  thought  it  was  sufficient  to  give  this  one,  since  he  was  to 
make  use  of  neither  of  them  in  his  mutilated  edition  of  the  5th 
book;  and  he  might  as  well  have  left  out  the  other,  as  also  the 
5th  proposition,  for  the  same  reason.  The  demonstration  of  the 
other  case  is  now  added,  because  both  of  them,  as  also  the  5th 
proposition,  are  necessary  to  the  demonstration  of  the  18th  propo- 
sition of  this  book.  The  translation  from  the  Arabic  gives  both 
cases  briefly. 

PROP.  A.  B.  V. 

This  proposition  is  frequently  used  by  geometers,  and  it  is  ne- 
cessary in  the  25th  prop,  of  this  book,  31st  of  the  6th,  and  34th 
of  the  11th,  and   15th  of  the    12th  book.     It  seems  to  have  been 
taken  out  of  the  Elements  by  Theon,  because  it  appeared  evident 
enough  to  him,  and  others,  who  substitute  the  confused  and  indis- 
tinct idea  the  vulgar  have  of^^ proportionals,  in  place  of  that  accu- 
rate idea  which  is  to  be  got  from  the  5th  definition  of  this  book. 
Nor  can  there  be  any  doubt  that  Eudoxus  or  Euclid  gave  it  a 
place  in  the  Elements,  when  we  see  the  7th  and  9th  of  the  same 
book  demonstrated,  though  they  are  quite  as  easy  and  evident  as 
this.     Alphonsus  Borellus  takes  occasion  from  this  proposition  to 
censure  the  5th  definition  of  this  book  very  severely,  but  most 
unjustly.     In  p.   126,  of  his  Euclid  restored,  printed  at  Pisa  in 
1658,  he  says,  "  Nor  can  even  this  least  degree  of  knowledge  be 
obtained  from  the  aforesaid  property,"  viz.  that  which  is  contained 
in  5th  def.  5.    "That  if  four  magnitudes  be  proportionals,  the 
third  must  necessarily  be  greater  than  the  fourth,  when  the  first 
is  greater  than  the  second;  as  Clavius  acknowledges  in  the  16th 
prop,  of  the  5th  book  of  the  Elements."     But  though  Clavius 
makes  no  such  acknowledgment  expressly,  he  has  given  Borellus 


BOOK.    V 


NOTES. 


255 


a  handle  to  say  this  of  him:  because  when  Clavius,  in  the  above 
cited  place,  censures  Commandiiie,  and  that  very  justly,  for  de- 
monstrating this  proposition  by  help  of  the  16th  of  the  fifth; 
yet  he  himself  gives  no  demonstration  of  it,  but  thinks  it  plain 
from  the  nature  of  proportionals,  as  he  writes  in  the  end  of  the 
14th  and  16th  prop,  book  5,  of  his  edition,  and  is  followed  by  He- 
rigon  in  Schol.  1,  prop.  14th,  book  5,  as  if  there  was  any  nature 
of  proportionals  antecedent  to  that  which  is  to  be  derived  and  un- 
derstood from  the  definition  of  them.  And,  indeed,  though  it  is 
very  easy  to  give  a  right  demonstration  of  it,  nobody,  as  far  as  I 
know,  has  given  one,  except  the  learned  Dr.  Barrow,  who,  in  an- 
swer to  Borellus's  objection,  demonstrates  it  indirectly,  but  very 
briefly  and  clearly,  from  the  5th  definition  in  the  322d  page  of  his 
Lect.  Mathem.  from  which  definition  it  may  also  be  easily  demon- 
strated directly.  On  which  account  we  have  placed  it  next  to  the 
propositions  concerning  equimultiples. 

PROP.  B.  B.  V. 

This  also  is  easily  deduced  from  the  5th  def.  b.  5,  and  there- 
fore is  placed  next  to  the  other;  for  it  was  very  ignorantly  made 
a  corollary  from  the  fourth  prop,  of  this  book.  See  the  note  on 
that  corollary.  ' 

PROP.  C,  B.  V. 

This  is  frequently  made  use  of  by  geometers,  and  is  necessary 
to  the  5th  and  6th  propositions  of  the  10th  book:  Clavius,  in 
his  notes  subjoined  to  the  8th  (\cf.  of  book  5,  demonstrates  it  only 
in  numbers,  by  help  of  some  of  the  propositions  of  the  7th  book; 
in  order  to  demonstrate  the  property  contained  in  the  5th  defini- 
tion of  the  5th  book,  when  applied  to  numbers,  from  the  property 
of  proportionals  contained  in  the  20th  def  of  the  7th  book.  And 
most  of  the  commentators  judge  it  difficult  to  prove  that  four 
magnitudes  which  are  proportionals  according  to  the  20th  def.  of 
7th  book,  are  also  proportionals  according  to  the  5th  def.  of  5th 
book.     But  this  is  easily  made  out,  as  follows: 

First,  If  A,  B,  C,  D,  be  four  mag- 
nitudes, such  that  A  is  the  same  multi- 
ple, or  the  same  part  of  B,  which  C  is 
of  D;  A,  B,  C,  D  are  proportionals. 
This  is  demonstrated  in  proposition  C. 

Secondly,  if  AB  contain  the  same 
parts  of  CD,  that  EF  does  of  GH;  in 


F 


B 


this  case  likewise  AB  is  to  CD,  as  EF 
to  GH. 


D 


K— 


C 


H 


L— 


E 


H 


Let  CK  be  a  part  of  CD,  and  GL  the  same  part  of  GH,  and 
let  AB  be  the  same  multiple  of  CK,  F 

that  EF  is  of  GL:  therefore,  by  prop.      B 
C  of  5th  book,  AB  is  to  CK,  as  EF  to  D 

GL:  and  CD,  GH  are  equimultiples 
of  CK,  GL  the  second  and  fourth: 
wherefore  by  cor.  prop.  4,  book  5,  AB 
is  to  CD,  as  EF  to  GH. 

And  if  four  magnitudes  be  propor- 
tionals according  to  the  5th  def.  of  book         A  C-       E  G 


K— 


256 


NOTES. 


BOOK  V, 


5,  they  are  also  proportionals    according   to    the   20th '  def.   of 

book  7.  " 

First,  if  A  be  to  B,  as  C  to  D;  then  if  A  be  any  multiple  or 
part  of  B,  C  is  the  same  multiple  or  part  of  D,  by  prop.  D,  of 
book  5. 

Next,  if  AB  be  to  CD,  as  EF  to  GH^  then  if  AB  contains  any 
parts  of  CD,  EF  contains  the  same  parts  of  GH:  for  let  CK  be  a 
part  of  CD,  and  GL  the  same  part  of  GH,  and  let  AB  be  a  mul- 
tiple of  CK,*  EF  is  the  same  multiple  of  GL;  take  M  the  same 
multiple  of  GL  that  AB  is  of  CK;  therefore  by  prop.  C,  of  book 
5,  AB  is  to  CK,  as  M  to  GL;  and  CD,  GH  are  equimultiples  of 
CK,  GL;  wherefore  by  cor.  prop.  4,  b.  5,  AB  is  to  CD,  as  M  to 
GH.  And,  by  the  hypothesis,  AB  is  to  CD  as  EF  to  GH:  there- 
fore M  is  equal  to  EF,  by  prop.  9,  book  5,  and  consequently  EF 
is  the  same  multiple  of  GL  that  AB  is  of  CK. 

PROP.  D.  B.  V. 

This  is  not  unfrequently  used  in  the  demonstration  of  other 
propositions,  and  is  necessary  in  that  of  prop.  9,  b.  6.  It  seems 
Theon  has  left  it  out  for  the  reasons  mentioned  in  the  notes  of 
prop. A. 

PROP.  vm.  B.  V. 

In  the  demonstration  of  this,  as  it  is  now  in  the  Greek,  there 
are  two  cases  (see  the  demonstration  in  Hervagius,  or  Dr.  Grego- 
ry's edition),  of  which  the  first  is  that  in  which  AE  is  less  than 
EB;  and  in  this  it  necessarily  follows,  that  H©  the  multiple  of 
EB  is  greater  than  ZH,  the  same  multiple  of  AE,  which  last 
multiple,  by  the  construction,  is  greater  than  A;  whence  also  H© 
must  be  greater  than  a.  But  in  the  second  case,  viz.  that  in  which 
EB  is  less  than  AE,  though  ZH  be  greater  than  A,  yet  H©  may 
be  less  than  the  same  A;  so  that  there  cannot  be  taken  a  multiple 
of  A  which  is  the  first  that  is  greater  than  K  or  H©  because  A  it- 
self is  greater  than  it;  upon  this  account  the  author  of  this  de- 
monstration found  it  necessary  to  change  one  part  of  the  construc- 
tion that  was  made  use  of  in  the  first  case:  but  he  has,  without 
any  necessity,  changed  also  another  part  of  it,  viz.  when  he  orders 
to  take  N  that  multiple  of  A  which 
is  the  first  that  is  greater  than  ZH; 
for  he  might  have  taken  that  mul- 
tiple of  A  which  is  the  first  that  is 
greater  than  H©,  or  K,  as  was 
done  in  the  first  case:  he  likewise  H-- 
brings  in  this  K  into  the  demon- 
stration of  both  cases,  v.'ithout  any 
reason;  for  it  serves  to  no  purpose 
but  to  lengthen  the  demonstration. 


E- 


0 


H— 


A 


13      A 


0 


I 


B   A 


There  is  also  a  third  case,  which 
is  not  mentioned  in  this  demon- 
stration, viz.  that  in  which  AE  in  the  first,  or  EB  in  the  second 
of  the  two  other  cases,  is  greater  than  D;  and  in  this  any  equi- 
multiples, as  the  doubles,  of  AE,  KB  are  to  be  taken,  as  is  done 
in  this  edition,  where  all  the  cases  are  at  once  demonstrated:  and 


BOOK  V.  NOTES.  257 

from  this  it  is  plain  that  Theon,  or  some  other  unskilful  editor 
has  vitiated  this  proposition. 

PROP.  IX.  B.  V. 

Of  this  there  is  given  a  more  explicit  demonstration  than  that 
which  is  now  in  the  Elements. 

PROP.  X.  B.  V. 

It  was  necessary  to  give  another  demonstration  of  this  propo- 
siiioi),  because  that  which  is  in  the  Greek  and  Latin,  or  other 
editions,  is  not  legitimate:  for  the  words  greater  ^the  same,  or  equal, 
lesser,  have  a  quite  different  meaning  when  applied  to  magnitudes 
and  ratios,  as  is  plain  from  the  5th  and  7th  definitions  of  book  5. 
By  the  help  of  these  let  us  examine  the  demonstration  of  the  10th 
prop,  which  proceeds  thus:  "  Let  A  have  to  C  a  greater  ratio  than 
B  to  C:  I  say  that  A  is  greater  than  B.     For  if  it  is  not  greater,  it 
it  is  either  equal,  or  less.     But  A  cannot  be  equal  to  B,  because 
then  each  of  them  would  have  the  same  ratio  to  C:  but  they  have 
not.     Therefore  A  is  not  equal  to  B."     The  force  of  which  rea- 
soning is  this;  if  A  had  to  C  the  same  ratio  that  B  has  to  C;  then 
if  any  equimultiples  whatever  of  A  and  B  be  taken,  and  any  mul- 
tiple whatever  of  C;  if  the  multiple  of  A  be  greater  than  the  multi- 
ple of  C,  then,  by  the  5th  def.  of  book  5,  the  multiple  of  B  is 
also  greater  than  that  of  C;  but,  from  the  hypothesis  that  A  has  a 
greater  ratio  to  C,  than  B  has  to  C,  there  must,  by  the  7th  def.  of 
book  5,  be  certain  equimultiples  of  A  and  B,  and  some  multiple 
of  C,  such  that  the  multiple  of  A  is  greater  than  the  multiple  of 
C,  but  the  multiple  of  B  is  not  greater  than  the  same  multiple  of 
C:  and  this  proposition  directly  contradicts  the  preceding:  where- 
fore A  is  not  equal  to  B.     The  demonstration  of  the  10th  prop, 
goes  on  thus:  "  but  neither  is  A  less  than  B;  because  then  A  would 
have  a  less  ratio  to  C  than  B  has  to  it:  but  it  has  not  a  less  ratio, 
therefore  A  is  not  less  than  B,"  8<:c.      Here  it  is  said,  that  "A 
would  have  a  less  ratio  to  C  than  B  has  to  C,"  or  which  is  the 
same  thing,  that  B  would  have  a  greater  ratio  to  C  than  A  to  C| 
that  is,  by  7th  def.  book  5,  there  must  be  some  equimultiples  of 
of  B  and  A,  and  some  multiple  of  C,  such  that  the  multiple  of  B 
is  greater  than  the  multiple  of  C,  but  the  multiple  of  A  is  not 
greater  than  it;  and  it  ought  to  have  been  proved,  that  this  can 
never  happen  if  the  ratio  of  A  to  C  be  greater  than  the  ratio  of 
B  to  C;  that  is,  it  should  have  been  proved,  that  in  this  case,  the 
multiple  of  A  is  always  greater  than  the  multiple  of  C,  whenever 
the  multiple  of  B  is  greater  than  the  multiple  of  C;  for  when  this 
is  demonstrated,  it  will  be  evident  that  B  cannot  have  a  greater 
ratio  to  C,  than  A  has  to  C,  or,  which  is  the  same  thing,  that  A 
cannot  have  a  less  ratio  to  C  than  B  has  to  C:  but  this  is  not  at  all 
proved  in  the  10th  proposition;  but  if  the  10th  were  once  demon- 
strated, it  would  immediately  follow  from  it,  but  cannot  without 
it  be  easily  demonstrated,  as  he   that  tries   to   do  it  will  find. 
Wherefore  the  10th  proposition  is  not  sufficiently  demonstrated. 
And  it  seems  that  he  who  has  given  the  demonstration  of  the  10th 
proposition  as  we  now  have  it,  instead  of  that  which  Eudoxus  or 
2  K 


258 


NOTES. 


BOOK  V. 


If  A  have  to  C   a 


A 
D 


C 

F 


B 
E 


C 
F 


Euclid  had  given,  has  been  deceived  in  applying  what  is  manifest, 
when  understood  of  magnitudes,  unto  ratios,  viz.  that  a  magnitude 
cannot  be  both  greater  and  less  than  another.  That  those  things 
which  are  equal  to  the  same  are  equal  to  one  another,  is  a  most 
evident  axiom  when  understood  of  magnitudes^  yet  Euclid  does 
not  make  use  of  it  to  infer  that  those  ratios  which  are  the  same  to 
the  same  ratio,  are  the  same  to  one  another^  but  explicitly  demon- 
strates this  in  prop.  11,  of  book  5.  The  demonstration  we  have 
given  of  the  10th  prop,  is  no  doubt  the  same  with  that  of  Eudoxus  or 
Euclid,  as  it  is  immediately  and  directly  derived  from  the  definition 
of  a  greater  ratio,  viz.  the  7th  of  the  5th. 

The  above  mentioned  proposition,  viz. 
greater  ratio  than  B  to  C;  and  if  of  A  and 
B  there  be  taken  certain  equimultiples,  and 
some  multiple  of  C;  then  if  the  multiple  of  B 
be  greater  than  the  multiple  of  C,  the  multi- 
ple of  A  is  also  greater  than  the  same,  is  thus 
demonstrated. 

Let  D,  E  be  equimultiples  of  A,  B,  and  F 
a  naultiple  of  C,  such,  that  E  the  multiple  of 
B  is  greater  than  F|  D  the  multiple  of  A  is 
also  greater  than  F. 

Because  A  as  a  greater  ratio  to  C,  than  B 
to  C,  A  is  greater  than  B,  by  the  10th  prop, 
book  5;  therefore  D  the  multiple  of  A  is 
greater  than  E  the  same  multiple  of  B:  and  E 
is  greater  than  F;  much  more  therefore  D  is 
greater  than  F. 

PROP.  XIII.  B.  V. 

In  Commandine's,  Briggs's,  and  Gregory's  translations,  at  the 
beginning  of  this  demonstration,  it  is  said,  "And  the  multiple  of 
C  is  greater  than  the  multiple  of  Dj  but  the  multiple  of  E  is  not 
greater  than  the  multiple  of  F;"  which  words  are  a  literal  transla- 
tion from  the  Greek;  but  the  sense  evidently  requires  that  it  be 
read,  "so  that  the  multiple  of  C  be  greater  than  the  multiple  of  D; 
but  the  multiple  of  E  be  not  greater  than  the  multiple  of  F."  And 
thus  this  place  was  restored  to  the  true  reading  in  the  first  editions 
of  Commandine's  Euclid,  printed  in  8vo.  at  Oxford;  but  in  the 
later  editions,  at  least  in  that  of  1747,  the  error  of  the  Greek  text 
was  kept  in. 

There  is  a  corollary  added  to  prop.  13,  as  it  is  necessary  to  the 
20th  and  21st  prop,  of  this  book,  and  is  as  useful  as  the  propo- 
sition. 

PROP.  XIV.  B.  V. 

The  two  cases  of  this,  which  are  not  in  the  Greek,  are  added; 
the  demonstration  of  them  not  being  exactly  the  same  with  that 
of  the  first  case. 

PROP.  XVII.  B.  V. 

The  order  of  the  words  in  a  clause  of  this  is  changed  to  one 
more  natural;  as  was  also  done  in  prop.  1. 


BOOK  V.  NOTES.  259 

PROP.  XVIII.  B.  V. 

The  demonstration  of  this  is  none  of  Euclid's,  nor  is  it  legiti- 
matej  for  it  depends  upon  this  hypothesis,  that  to  any  three  mag- 
nitudes, two  of  which,  at  least,  are  of  the  same  kind,  there  may 
be  a  fourth  proportional:  which,  if  not  proved,  the  demonstration 
now  in  the  text  is  of  no  force:  but  this  is  assumed  without  any 
proofs  nor  can  it  as  far  as  I  am  able  to  discern,  be  demonstrated 
by  the  propositions  preceding  this:  so  far  is  it  from  deserving  to 
be  reckoned   an   axiom,  as   Clavius,  after  other   commentators, 
would   have  it,  at  the  end   of  the   definitions  of  the   5th  book. 
Euclid  does  not  demonstrate  it,  nor  does  he  show  how  to  find  the 
fourth  proportional,  before  the  12th  prop,  of  the  sixth  book:  and 
he  never  assumes  any  thing  in   the  demonstration  of  a  proposi- 
tion, which  he  had  not  before  demonstrated:  at  least,  he  assumes 
nothing  the  existence  of  which  is  not  evidently  possible;  for  a 
certain  conclusion  can  never  be  deduced  by  the  means  of  an  un- 
certain proposition:  upon  this  account,  we  have  given  a  legitimate 
demonstration  of  this  proposition  instead  of  that  in  the  Greek 
and  other  editions,  which  very  probably  Theon,  at  least  some 
other,  has  put  in  the  place  of  Euclid's,  because  he  thought  it  too 
prolix:  and  as  the  17th  prop,  of  which  this  18th  is  the  converse, 
is  demonstrated  by  help  of  the  1st  and  second  propositions  of  this 
book;  so,  in  the  demonstration  now  given  of  the  18th,  the  5th  prop, 
and  both  cases  of  the  6th  are  necessary,  and  these  two  propo- 
sitions are  the  converses  of  the  1st  and  2d.     Now  the  5th  and  6th 
do  not  enter  into  the  demonstration  of  any  proposition  in  this 
book  as  we  now  have  it:  nor  can  they  be  of  use  in  any  proposi- 
tion of  the  Elements,  except  in  this  18th,  and  this  is  a  manifest  proof, 
that  Euclid  made  use  of  them  in  his  demonstration  of  it,  and  that 
the  demonstration  now  given,  which  is  exactly  the  converse  of 
that  of  the  17th,  as  it  ought  to  be,  differs  nothing  from  that  of 
Eudoxus  or  Euclid:  for  the  5th  and  6th  have  undoubtedly  been 
put  into  the  5th  book  for  the  sake  of  some  propositions  in  it,  as 
all  the  other  propositions  about  equimultiples  have  been. 

Hieronymus  Saccherius,  in  his  book  named  Euclides  ab  omni 
naevo  vindicatus,  printed  at  Milan,  anno  1733,  in  4to,  acknowledges 
this  blemish  in  the  demonstration  of  the  18th,  and  that  he  may  re- 
move it,  and  render  the  demonstration  we  now  have  of  it  legiti- 
mate, he  endeavours  to  demonstrate  the  following  proposition, 
which  is  in  page  II  5  of  his  book,  viz. 

"  Let  A,  B,  C,  D  be  four  magnitudes,  of  which  the  two  first  are 
,  of  the  one  kind,  and  also  the  two  others  either  of  the  same  kind 
with  the  two  first,  or  of  some  other,  the  same  kind  with  one  ano- 
ther. I  say  the  ratio  of  the  third  C  to  the  fourth  D,  is  either 
equal  to,  or  greater,  or  less  than  the  ratio  of  the  first  A  to  the 
second  B." 

And  after  two  propositions  premised  as  lemmas,  he  proceeds 
thus: 

"  Either  among  all  the  possible  equimultiples  of  the  first  A,  and 
of  the  third  C,  and  at  the  same  time,  among  all  the  possible  equi- 
multiples of  the  second  B,  and  of  the  fourth  D,  there  can  be  found 


260  NOTES  BOOK  V; 

some  one  multiple  EF  of  the  first  A,  and  one  IK  of  the  second  B, 
that  are  equal  to  one  another;  and  also,  in  the  same  case,  some 
one  multiple  GH  of  the  third  C  equal  to  LM  the  multiple  of  the 
fourth  D,  or  such  equality  is  no  where  to  be  found.  If  the  first 
case  happen  [i.  e.  if 

such  equality  is   to     A E F 

be  found]  it  is  mani- 
fest from  what  is  be-     B I K 

fore    demonstrated, 

that  A  is  to  B,  as  C     C G H 

to  D;  but  if  such  si-  ' 

multaneous  equality     D E M 

be  not  to  be  found 

upon  both  sides,  it  will  be  found  either  upon  one  side,  as  upon  the 
side  of  A  [and  B;]  or  it  will  be  found  upon  neither  side;  if  the 
first  happen;  therefore  (from  Euclid's  definition  of  greater  and 
lesser  ratio  foregoing)  A  has  to  B  a  greater  or  less  ratio  than  C 
to  D;  according  as  GH  the  multiple  of  the  third  C  is  less,  or 
greater  than  LM  the  multiple  of  the  fourth  D:  but  if  the  second 
case  happen;  therefore  upon  the  one  side,  as  upon  the  side  of  A 
the  first  and  B  the  second,  it  may  happen  that  the  multiple  EF, 
[viz.  of  the  first]  may  be  less  than  IK  the  multiple  of  the  second, 
while,  on  the  contrary,  upon  the  other  side,  [viz.  of  C  and  D]  the 
multiple  GH  [of  the  third  C]  is  greater  than  the  other  multiple 
LM  [of  the  fourth  D:]  and  then  (from  the  same  definition  of  Eu- 
clid) the  ratio  of  the  first  A  to  the  second  B,  is  less  than  the  ratio 
of  the  third  C  to  the  fourth  D;  or  on  the  contrary. 

"  Therefore  the  axiom  [i.  e.  the  proposition  before  set  down] 
remains  demonstrated,"  Sec. 

Not  in  the  least;  but  it  remains  still  undemonstrated;  for  what 
he  says  may  happen,  may,  in  innumerable  cases,  never  happen; 
and  therefore  his  demonstration  does  not  hold:  for  example,  if  A 
be  the  side,  and  B  the  diameter  of  a  square;  and  C  the  side,  and 
D  the  diameter  of  another  square;  there  can  in  no  case  be  any 
multiple  of  A  equal  to  any  of  B;  nor  any  one  of  C  equal  to  one 
of  D,  as  is  well  known;  and  yet  it  can  never  happen,  that  when 
any  multiple  of  A  is  greater  than  a  multiple  of  B,  the  multiple  of 
C  can  be  less  than  the  multiple  of  D,  nor  when  the  multiple  of  A 
is  less  than  that  of  B,  the  multiple  of  C  can  be  greater  than  that 
of  D,  viz.  taking  equimultiples  of  A  and  C,  and  equimultiples  of 
B  and  D:  for  A,  B,  C,  D  are  proportionals;  and  so  if  the  multiple 
of  A  be  greater,  Sec.  than  that  of  B,  so  must  that  of  C  be  greater, 
&c.  than  that  of  D;  by  5th  def.  b.  5. 

The  same  objection  holds  good  against  the  demonstration  which 
some  give  of  the  1st  prop,  of  the  6th  book,  which  we  have  made 
against  this  of  the  18th  prop,  because  it  depends  upon  the  same 
insufficient  foundation  with  the  other. 

PROP.  XIX.  B.  V. 

A  corollary  is  added  to  this,  which  is  as  frequently  used  as  the 
proposition  itself.  The  corollary  which  is  subjoined  to  it  in  the 
Greek,  plainly  shows  tliat  the  5th  book  has  been  vitiated  by  edi- 


BOOK  V.  NOTES.  261 

tors  who  were  not  geometers:  for  the  conversion  of  ratios  does 
not  depend  upon  this  19th,  and  the  demonstration  which  several 
of  the  commentators  on  Euclid  give  of  conversion  is  not  legiti- 
mate, as  Clavius  has  rightly  observed,  who  has  given  a  good  de- 
monstration of  it,  which  we  have  put  in  proposition  E;  but  he 
makes  it  a  corollary  from  the  19th,  and  begins  it  with  the  words, 
"Hence  it  easily  follows,"  though  it  does  not  at  all  follow  from  it. 

PROP.  XX.  XXI.  XXII.  XXIII.  XXIV.   B.  V. 

The  demonstrations  of  the  20th  and  21st  propositions,  are 
shorter  than  those  Euclid  gives  of  easier  propositions,  either  in 
the  preceding  or  following  books:  wherefore  it  was  proper  to 
make  them  more  explicit,  and  the  22d  and  23d  propositions  are, 
as  they  ought  to  be,  extended  to  any  number  of  magnitudes:  and, 
in  like  manner  may  the  24th  be,  as  is  taken  notice  of  in  a  corol- 
lary^ and  another  corollary  is  added,  as  useful  as  the  proposition, 
and  the  words  "any  whatever"  are  supplied  near  the  end  of  prop. 
23,  which  are  wanting  in  the  Greek  text,  and  the  translations 
from  it. 

In  a  paper  writ  by  Philippus  Naudaeus,  and  published  after  his 
death,  in  the  History  of  the  Royal  Academy  of  Sciences  of  Berlin, 
anno  1745,  page  50,  the  23d  prop,  of  the  5th  book  is  censured  as 
being  obscurely  enunciated,  and,  because  of  this,  prolixly  demon- 
strated: the  enunciation  there  given  is  not  Euclid's,  but  Tacquet's, 
as  he  acknowledges,  which,  though  not  so  well  expressed,  is,  upon 
the  matter,  the  same  with  that  which  is  now  in  the  Elements. 
Nor  is  there  any  thing  obscure  in  it,  though  the  author  of  the  paper 
has  set  down  the  proportionals  in  a  disadvantageous  order,  by 
which  it  appears  to  be  obscure:  but,  no  doubt,  Euclid  enunciated 
this  23d,  as  well  as  the  22d,  so  as  to  extend  it  to  any  number  of 
magnitudes,  which  taken  two  and  two  are  proportionals,  and  not 
of  six  onlyj  and  to  this  general  case  the  enunciation  which  Nau- 
daeus gives,  cannot  be  well  applied. 

The  demonstration  which  is  given  of  this  23d,  in  that  paper,  is 
quite  wrong;  because,  if  the  proportional  magnitudes  be  plane  or 
solid  figures,  there  can  no  rectangle  (which  he  improperly  calls  a 
product)  be  conceived  to  be  made  by  any  two  of  them,  and  if  it 
should  be  said  that  in  this  case  straight  lines  are  to  be  taken 
which  are  proportional  to  the  figures,  the  demonstration  would 
this  way  become  much  longer  than  Euclid's:  but,  even  though  his 
demonstration  had  been  right,  who  does  not  see  that  it  could  not 
be  made  use  of  in  the  5th  book.^ 

PROP.  F,  G,  H,  K.    B.  V. 

These  propositions  are  annexed  to  the  5th  book,  because  they 
are  frequently  made  use  of  by  both  ancient  and  modern  geometers: 
and  in  many  cases  compound  ratios  cannot  be  brought  into  de- 
monstration, without  making  use  of  them. 

Whoever  desires  to  see  the  doctrine  of  ratios  delivered  in  this 
5th  book  solidly  defended,  and  the  arguments  brought  against  it 
by  And.  Tacquet,  Alph.  Rorellus,  and  others,  fully  refuted,  may 


262  NOTES.  BOOK  VI. 

read  Dr.  Barrow's  Mathematical  Lectures,  viz.  the  7th  and  8th  of 
the  year  1666. 

'  The  5th  book  being  thus  corrected,  I  most  readily  agree  to 
what  the  learned  Dr.  Barrow  says,*  "  That  there  is  nothing  in  the 
whole  body  of  the  Elements  of  a  more  subtile  invention,  nothing 
more  solidly  established,  and  more  accurately  handled,  than  the 
doctrine  of  proportionals."  And  there  is  some  ground  to  hope, 
that  georneters  will  think  that  this  could  not  have  been  said  with 
as  good  reason,  since  Theon's  time  till  the  present. 

DEF.  II.  and  V.  of  B.  VI. 

The  2d  definition  does  not  seem  to  be  Euclid's,  but  some  un- 
skilful editor's:  for  there  is  no  mention  made  by  Euclid,  nor,  as 
far  as  I  know,  by  any  other  geometer,  of  reciprocal  figures:  it  is 
obscurely  expressed,  which  made  it  proper  to  render  it  more  dis- 
tinct: it  would  be  better  to  put  the  following  definition  in  place  of 
it,  viz. 

DEF.  II. 

Two  magnitudes  are  said  to  be  reciprocally  proportional  to  two 
others,  when  one  of  the  first  is  to  one  of  the  other  magnitudes,  as 
the  remaining  one  of  the  last  two  is  to  the  remaining  one  of 
the  first. 

But  the  fifth  definition,  which,  since  Theon's  time,  has  been  kept 
in  the  Elements,  to  the  great  detriment  of  learners,  is  now  justly 
thrown  out  of  them,  for  the  reason  given  in  the  notes  on  the  23d 
prop,  of  this  book. 

PROP.  I.  and  II.    B.  VI. 

To  the  first  of  these  a  corollary  is  added,  which  is  often  used: 
and  the  enunciation  of  the  second  is  made  more  general. 

PROP.  III.   B.  VI. 

A  second  case  of  this,  as  useful  as  the  first,  is  given  in  prop.  A: 
viz.  the  case  in  which  the  exterior  angle  of  a  triangle  is  bisected 
by  a  straight  line:  the  demonstration  of  it  is  very  like  to  that  of 
the  first  case,  and  upon  this  account  may,  probably,  have  been 
left  out,  as  also  the  enunciation,  by  some  unskilful  editor:  at  least, 
it  is  certain,  that  Pappus  makes  use  of  this  case  as  an  elementary 
proposition,  without  a  demonstration  of  it,  in  prop.  39  of  his  7th 
book  of  Mathematical  Collections. 

PROP.  VII.     B.  VI. 

To  this  a  case  is  added  which  occurs  not  unfrequently  in  de- 
monstration. 

PROP.  VIII.     B.  VI. 

It  seems  plain  that  some  editor  has  changed  the  demonstration 
that  Euclid  gave  of  this  proposition:  for,  after  he  has  demon- 
strated, that  the  triangles  are  equiangular  to  one  another,  he  par- 
ticularly shows  that  their  sides  about  the  equal  angles  are  pro- 

*  See  page  336. 


BOOK  VI.  NOTES.  263 

portionals,  as  if  this  had  not  been  done  in  the  demonstration  of 
the  4th  prop,  of  this  book;  this  superfluous  part  is  not  found  in 
the  translation  from  the  Arabic,  and  is  now  left  out. 

PROP.  IX.     B.  VI. 

This  is  demonstrated  in  a  particular  case,  viz.  that  in  which 
the  third  part  of  a  straight  line  is  required  to  be  cut  off;  which  is 
not  at  all  like  Euclid's  manner:  besides,  the  author  of  the  demon- 
stration, from  four  magnitudes  being  proportionals,  concludes 
that  the  third  of  them  is  the  same  multiple  of  the  fourth,  which 
the  first  is  of  the  second:  now,  this  is  no  where  demonstrated  in 
the  5th  book,  as  we  now  have  it:  but  the  editor  assumes  it  from 
the  confused  notion  which  the  vulgar  have  of  proportionals:  on 
this  account,  it  was  necessary  to  give  a  general  and  legitimate  de- 
monstration of  this  proposition. 

PROP.  XVIII.     B.  VI. 

The  demonstration  of  this  seems  to  be  vitiated:  for  the  propo- 
sition is  demonstrated  only  in  the  case  of  quadrilateral  figures, 
without  mentioning  how  it  may  be  extended  to  figures  of  five  or 
more  sides:  besides,  from  two  triangles  being  equiangular,  it  is 
inferred  that  a  side  of  the  one  is  to  the  homologous  side  of  the 
other,  as  another  side  of  the  first  is  to  the  side  homologous  to  it 
of  the  other,  without  permutation  of  the  proportionals;  which  is 
contrary  to  Euclid's  manner,  as  is  clear  from  the  next  proposi- 
tion; and  the  same  fault  occurs  again  in  the  conclusion,  where 
the  sides  about  the  equal  angles  are  not  shown  to  be  proportion- 
als, by  reason  of  again  neglecting  permutation.  On  these  ac- 
counts, a  demonstration  is  given  in  Euclid's  manner,  like  to  that 
he  makes  use  of  in  the  20th  prop,  of  this  book :  and  it  is  extended 
to  five-sided  figures,  by  which  it  may  be  seen  how  to  extend  it  to 
figures  of  any  number  of  sides. 

PROP.  XXIII.     B.  VI. 

Nothing  is  usually  reckoned  more  difficult  in  the  Elements  of 
geometry  by  learners,  than  the  doctrine  of  compound  ratio,  which 
Theon  has  rendered  absurd  and  ungeometrical,  by  substituting 
the  5th  definition  of  the  6th  book  in  place  of  the  right  definition, 
which  without  doubt  Eudoxus  or  Euclid  gave,  in  its  proper  place, 
after  the  definition  of  triplicate  ratio,  Sec.  in  the  5th  book.  The- 
on's  definition  is  this:  a  ratio  is  said  to  be  compounded  of  ratios 

oTotv  (X.I  Tav  Mya^  TTjjA/xoT^jTes  tCp*  ictvrctc,  TroXXxTrXxcrtocT^eio-cci  Trotcacri  riyati 
which  Commandine  thus  translates;  "quando  rationem  quanti- 
tates  inter  se  multiplicatae  aliquam  efficiunt  rationem;"  that  is, 
when  the  quantities  of  the  ratios  being  multiplied  by  one  another 
make  a  certain  ratio.  Dr.  Wallis  translates  the  word  7r7}Xiy.aTt3r£(; 
"rationem  exponentes,"  the  exponents  of  the  ratios:  and  Dr.  Gre- 
gory renders  the  last  words  of  the  definition  by  "illius  facit  quan- 
titatem,"  makes  the  quantity  of  that  ratio;  but  in  whatever  sense 
the  "quantities,"  or  "exponents  of  the  ratios,"  and  their  "multi- 
plication  be  taken,  the  definition  will  be  ungeometrical  and  use- 
less: for  there  can  be  no  multiplication  but  by  a  number.     Now 


V 


264  NOTES.  BOOK  VI. 

the  quantity  or  exponent  of  a  ratio  (according  as  Eutochius  in 
his  Comment,  on  prop.  4,  book  2,  of  Arch,  de  Sph.  et  Cyl.  and 
the  moderns  explain  that  term)  is  the  number  which  multiplied 
into  the  consequent  term  of  a  ratio  produces  the  antecedent,  or 
which  is  the  same  thing,  the  number  which  arises  by  dividing  the 
antecedent  by  the  consequent;  but  there  are  many  ratios  such, 
that  no  number  can  arise  from  the  division  of  the  antecedent  by 
the  consequent:  ex.  gr.  the  ratio  which  the  diameter  of  a  square 
has  to  the  side  of  it;  and  the  ratio  which  the  circumference  of  a 
circle  has  to  its  diameter,  and  such  like.  Besides,  that  there  is 
not  the  least  mention  made  of  this  definition  in  the  writings  of 
Euclid,  Archimedes,  Apollonius,  or  other  ancients,  though  they 
frequently  make  use  of  compound  ratio;  and  in  this  23d  prop,  of 
the  6th  book,  where  compound  ratio  is  first  mentioned,  there  is 
not  one  word  which  can  relate  to  this  definition,  though  here,  if 
in  any  place,  it  was  necessary  to  be  brought  in;  but  the  right  de- 
finition is  expressly  cited  in  these  words:  "But  the  ratio  of  K  to 
M  is  compounded  of  the  ratio  of  K  to  L,  and  of  the  ratio  of  L  to 
M."  This  definition  therefore  of  Theon  is  quite  useless  and  ab- 
surd: for  that  Theon  brought  it  into  the  Elements  can  scarce  be 
doubted;  as  it  is  to  be  found  in  his  commentary  upon  Ptolemy's 
M£ycc>i7}  'ZvvTu^tgj  page  62,  where  he  also  gives  a  childish  explica- 
tion of  it,  as  agreeing  only  to  such  ratios  as  can  be  expressed  by 
numbers;  and  from  this  place  the  definition  and  explication  have 
been  exactly  copied  and  prefixed  to  the  definitions  of  the  6th  book, 
as  appears  from  Hervagius's  edition:  but  Zambertus  and  Com- 
mandine,  in  their  Latin  translations,  subjoin  the  same  to  these  de- 
finitions. Neither  Campanus,  nor,  as  it  seems,  the  Arabic  manu- 
scripts, from  which  he  made  his  translation,  have  this  definition. 
Clavius,  in  his  observations  upon  it,  rightly  judges,  that  the  defi- 
nition of  compound  ratio  might  have  been  made  after  the  same 
manner  in  which  the  definitions  of  duplicate  and  triplicate  ratio 
are  given;  viz.  "  That  as  in  several  magnitudes  that  are  continual 
proportionals,  Euclid  named  the  ratio  of  the  first  to  the  third,  the 
duplicate  ratio  of  the  first  to  the  second,  and  the  ratio  of  the  first 
to  the  fourth,  the  triplicate  ratio  of  the  first  to  the  second,  that  is, 
the  ratio  compounded  of  two  or  three  intermediate  ratios  that  are 
equal  to  one  another,  and  so  on;  so,  in  like  manner,  if  there  be 
several  magnitudes  of  the  same  kind,  following  one  another, 
which  are  not  continual  proportionals,  the  first  is  said  to  have  to 
the  last  the  ratio  compounded  of  all  the  intermediate  ratios — only 
for  this  reason,  that  these  intermediate  ratios  are  interposed  be- 
twixt the  two  extremes,  viz.  the  first  and  last  magnitudes;  even 
as,  in  the  10th  definition  of  the  5th  book,  the  ratio  of  the  first  to 
the  third  was  called  the  duplicate  ratio,  merely  upon  account  of 
two  ratios  being  interposed  betv/ixt  the  extremes,  that  are  equal 
to  one  another:  so  that  there  is  no  difference  betwixt  this  com- 
pounding of  ratios,  and  the  duplication  or  triplication  of  them 
which  are  defined  in  the  5th  book,  but  that  in  the  duplication,  tri- 
plication, 8cc.  of  ratios,  all  the  interposed  ratios  are  equal  to  one 
another;  whereas,  in  the  compounding  of  ratios,  it  is  not  neces- 
sary that  the  intermediate  ratios  should  be  equal  to  one  another." 


BOOK  VI.  NOTES.  265 

Also  Mr.  Edmund  Scarburgh,  in  his  English  translation  of  the 
first  six  books,  page  238,  266,  expressly  affirms,  that  the  5th  defi- 
nition of  the  6th  book  is  supposititious,  and  that  the  true  de.fini^ 
tion  of  compound  ratio  is  contained  in  the  10th  definition  of  the 
5th  book,  viz.  the  definition  of  duplicate  ratio,  or  to  be  understood 
from  it,  to  wit,  in  the  same  manner  as  Clavius  has  explained  it  in 
the  preceding  citation.  Yet  these,  and  the  rest  of  the  moderns, 
do  notwithstanding  retain  this  5th  def.  of  the  6th  book,  and  illus- 
trate and  explain  it  by  long  commentaries,  when  they  ought  rather 
to  have  taken  it  quite  away  from  the  Elements, 

For,  by  comparing  def.  5,  book  6,  with  prop.  5,  book  8,  it  will 
clearly  appear  that  this  definition  has  been  put  into  the  Elements 
in  place  of  the  right  one,  which  has  been  taken  out  of  them:  be- 
cause, in  prop.  5,  book  8,  it  is  demonstrated  that  the  plane  num- 
ber of  which  the  sides  are  C,  D  has  to  the  plane  number  of  which 
the  sides  are  E,  Z  (see  Hervagius's  or  Gregory's  edition,)  the 
ratio  which  is  compounded  of  the  ratios  of  their  sides;  that  is,  of 
the  ratios  of  C  to  E,  and  D  to  Z :  and,  by  def.  5,  book  6,  and  the 
explication  given  of  it  by  all  the  commentators,  the  ratio  which 
is  compounded  of  the  ratios  of  C  to  E,  and  D  to  Z,  is  the  ratio 
of  the  product  made  by  the  multiplication  of  the  antecedents, 
C,  D,  to  the  product  by  the  consequents  E,  Z,  that  is,  the  ratio 
of  the  plane  number  of  which  the  sides  are  C,  D  to  the  plane 
number  of  which  the  sides  are  E,  Z.  Wherefore  the  proposi- 
tion which  is  the  5th  def.  of  book  6,  is  the  very  same  with  the 
5th  prop,  of  book  8,  and  therefore  it  ought  necessarily  to  be  can- 
celled in  one  of  these  places:  because  it  is  absurd  that  the  same 
proposition  should  stand  as  a  definition  in  one  place  of  the  Ele- 
ments, and  be  demonstrated  in  another  place  of  them.  Now, 
there  is  no  doubt  that  prop.  5,  book  8,  should  have  a  place  in  the 
Elements,  as  the  same  thing  is  demonstrated  in  it  concerning  plane 
numbers,  which  is  demonstrated  in  prop.  23,  book  6,  of  equiangular 
parallelograms;  wherefore  def.  5,  book  6,  ought  not  to  be  in  the 
Elements.  And  from  this  it  is  evident  that  this  definition  is  not 
Euclid's,  but  Theon's,  or  some  other  unskilful  geometer's. 

But  nobody,  as  far  as  I  know,  has  hitherto  shown  the  true  use 
of  compound  ratio,  or  for  what  purpose  it  has  been  introduced 
into  geometry:  for  every  proposition  in  which  compound  ratio  is 
made  use  of,  may  without  it  be  both  enunciated  and  demonstrated. 
Now  the  use  of  compound  ratio  consists  wholly  in  this,  that  by 
means  of  it  circumlocutions  may  be  avoided,  and  thereby  propo- 
sitions may  be  more  briefly  either  enunciated  or  demonstrated,  or 
both  may  be  done:  for  instance,  if  this  23d  proposition  of  the 
sixth  book  were  to  be  enunciated,  without  mentioning  compound 
ratio,  it  might  be  done  as  follows.  If  two  parallelograms  be  equi- 
angular, and  if  as  a  side  of  the  first  to  a  side  of  the  second,  so  any 
assumed  straight  line  be  made  to  a  second  straight  line:  and  as 
the  other  side  of  the  first  to  the  other  side  of  the  second,  so  the 
second  straight  line  be  made  a  third.  The  first  parallelogram  is 
to  the  second,  as  the  first  straight  line  to  the  third.  And  the  de- 
monstration would  be  exactly  the  same  as  we  now  have  it.  But 
the  ancient  geometers,  when  they  observed  this  enunciation  could 

2L 


266  NOTES.  BOOK  VI. 

be  made  shorter,  by  giving  a  name  to  the  ratio  which  the  first 
straight  line  has  to  the  last,  by  which  name  the  intermediate  ratios 
might  likewise  be  signified,  of  the  first  to  the  second,  and  of  the 
second  to  the  third,  and  so  on,  if  there  were  more  of  them,  they 
called  this  ratio  of  the  first  to  the  last  the  ratio  compounded  of  the 
ratios  of  the  first  to  the  second,  and  of  the  second  to  the  third 
straight  line:  that  is,  in  the  present  example,  of  the  ratios  which 
are  the  same  with  the  ratios  of  the  sides,  and  by  this  they  ex- 
pressed the  proposition  more  briefly,  thus:  if  there  be  two  equi- 
angular parallelograms,  they  have  to  one  another  the  ratio  which 
is  the  same  witii  that  which  is  compounded  of  ratios  that  are  the 
same  with  the  ratios  of  the  sides.  Which  is  shorter  than  the  pre- 
ceding enunciation,  but  has  precisely  the  same  meaning.  Or  yet 
shorter  thus:  equiangular  parallelograms  have  to  one  another  the 
ratio  which  is  the  same  with  that  which  is  compounded  of  the 
ratios  of  their  sides.  And  these  two  enunciations,  the  first  espe- 
cially, agree  to  the  demonstration  which  is  now  in  the  Greek. 
The  proposition  may  be  more  briefly  demonstrated,  as  Candalla 
does,  thus:  let  ABCD,  CEFG,  be  two  equiangular  parallelograms, 
and  complete  the  parallelogram  CDHG,  then,  because  there  are 
three  parallelograms  AC,  CH,  CF,  the  first  AC  (by  the  definition 
of  compound  ratio)  has  to  the  third  CF,  the  ratio  which  is  com- 
pounded of  the  ratio  of  the  first  AC  to  the  A  -^  H 
second  CH,  and  of  the  ratio  of  CH  to  the 
third  CF;  but  the  parallelogram  AC  is  to 

the  parallelogram  CH,  as  the  straight  line     q 

BC  to  CG;  and  the  parallelogram  CH  is  to     ^  '^ 

CF,  as  the  straight  line  CD  is  to  CE:  there- 
fore the  parallelogram  AC  has  to  CF  the 
ratio  which  is  compounded  of  ratios  that  -^     ^ 

are  the  same  with  the  ratios  of  the  sides.  And  to  this  demon- 
stration agrees  the  enunciation  which  is  at  present  in  the  text,  viz. 
equiangular  parallelograms  have  to  one  another  the  ratio  which 
is  compounded  of  the  ratios  of  the  sides;*  for  the  vulgar  reading, 
"  which  is  compounded  of  their  sides,"  is  absurd.  But  in  this 
edition,  we  have  kept  the  demonstration  which  is  in  the  Greek 
text,  though  not  so  short  as  Candalla's;  because  the  way  of  finding 
the  ratio  which  is  compounded  of  the  ratios  of  the  sides,  that  is, 
of  finding  the  ratio  of  the  parallelograms,  is  shown  in  that,  but  not 
in  Candalla's  demonstration;  whereby  beginners  may  learn,  in  like 
cases,  how  to  find  the  ratio  which  is  compounded  of  two  or  more 
given  ratios.  ♦ 

From  what  has  been  said,  it  may  be  observed,  that  in  any  mag- 
nitudes whatever  of  the  same  kind  A,  B,  C,  D,  Sec.  the  ratio  com- 
pounded of  the  ratios  of  the  first  to  the  second,  of  the  second  to 
the  third,  and  so  on  to  the  last,  is  only  a  name  or  expression,  by 
which  the  ratio  which  the  first  A  has  to  tl»e  last  D  is  signified,  and 
by  which  at  the  same  time  the  ratios  of  all  the  magnitudes  A 
to  B,  B  to  C,  C  to  D,  from  the  first  to  the  last,  to  one  another, 
whether  they  be  the  same,  or  be  not  the  same,  arc  indicated;  as  in 
magnitudes  which  are  continual  proportionals  A,  B,  C,  D,  &c.  the 
duplicate  'ratio  of  the  first  to  the  second  is  only  a  name  or  ex- 


B              C 

BOOK    VI.  NOTES.  267 

pression  by  which  the  ratio  of  the  first  A  to  the  third  C  is  signified, 
and  by  which,  at  the  same  time,  is  shown  that  there  are  two  ratios 
of  the  magnitudes,  from  the  first  to  the  last,  viz.  of  the  first  A  to 
the  second  B,  and  of  the  second  B  to  the  third  or  last  C,  which 
are  the  same  with  one  another;  and  the  triplicate  ratio  of  the  first 
to  the  second  is  a  name  or  expression  by  which  the  ratio  of  the 
first  A  to  the  fourth  D  is  signified,  and  by  which,  at  the  same  time, 
is  shown  that  there  are  three  ratios  of  the  magnitudes,  from  the 
first  to  the  last,  viz.  of  the  first  A  to  the  second  B,  and  of  B  to  the 
third  C,  and  of  C  to  the  fourth  or  last  D,  which  are  all  the  same 
with  one  another;  and  so  in  the  case  of  any  other  multiplicate  ratios. 
And  that  this  is  the  right  explication  of  the  meaning  of  these  ratios 
is   plain  from  the  definitions  of  duplicate  and  triplicate  ratio,  in 
which  Euclid  makes  use  of  the  word  ^^eyerui^  is  said  to  be,  or  is 
called,  which  word,  he,  no  doubt,  made  use  of  also  in  the  definition 
of  compound   ratio,  which  Theon,  or  some  other,  has  expunged 
from  the  Elements;  for  the  very  same  word  is  still  retained  in  the 
wrong  definition  of  compound  ratio,  which  is  now  the  5th  of  the 
6th  book:  but  in  the  citation  of  these  definitions  it  is  sometimes 
retained,  as  in  the  demonstration  of  prop.  19,  book  6,  "the  first 
is  said  to  have,  ex^iv  Xiyercit^  to  the  third  the  duplicate  ratio,"  Sec. 
which  is  wrong  translated  by  Commandine  and  others,  "has"  in- 
stead of  "is  said  to  have:"  and  sometimes  it  is  left  out,  as  in  the 
demonstration  of  prop.  33,  of  the  eleventh  book,  in  which  we  find 
"  the  first  has,  ex.si,  to  the  third  the  triplicate  ratio;"  but  without 
doubt  ex^i-,  "  has"  in  this  place  signifies  the  same  as  eptJ^'"  AcyfTee/, 
is  said  to  have:  so  likewise  in  prop.  23,  B.  6,  we  find  this  citation, 
"  but  the  ratio  of  K  to  M  is  compounded,  G-vyKurcct  of  the  ratio  of 
K  to  L,  and  the  ratio  of  L  to  M,"  which  is  a  shorter  way  of  ex- 
pressing the  same  thing,  which,  according  to  the  definition,  ought 
to  have  been  expressed  by  o-vyKna-^ui  Xeyercci^  is  said  to  be  com- 
pounded. 

From  these  remarks,  together  with  the  proposition  subjoined, 
to  the  5th  book,  all  that  is  found  concerning  compound  ratio, 
cither  in  the  ancient  or  modern  geometers,  may  be  understood 
and  explained. 

PROP.  XXIV.    B.  VI. 

It  seems  that  some  unskilful  editor  has  made  up  this  demon- 
stration as  we  now  have  it,  out  of  two  others;  one  of  which  may 
be  made  from  the  2d  prop,  and  the  other  from  the  4th  of  this 
book:  for  after  he  has,  from  the  2d  of  this  book,  and  composi- 
tion and  permutation,  demonstrated,  that  tlie  sides  about  the 
angle  common  to  the  two  parallelograms  are  proportionals,  he 
might  have  immediately  concluded,  that  the  sides  about  the  other 
equal  angles  were  proportionals,  viz.  from  prop.  34,  B.  1,  and 
prop.  7,  book  5.  This  he  does  not,  but  proceeds  to  show,  that 
the  triangles  and  parallelograms  are  equiangular;  and  in  a  tedious 
way  by  help  of  prop.  4,  of  this  book,  and  the  22d  of  book  5,  de- 
duces the  same  conclusion:  from  which  it  is  plain  that  this  ill 
composed  demonstration  is  not  Euclid's:  these  superfluous  things 
are  now  left  out,  and  a  more  simple  demonstration  is  given  from 


268  NOTES.  BOOK    VI. 

the  fourth  prop,  of  this  book,  the  same  which  is  in  the  trans- 
lation from  the  Arabic,  by  help  of  the  2d  prop,  and  composition: 
but  in  this  the  author  neglects  permutation,  and  does  not  show 
the  parallelograms  to  be  equiangular,  ias  is  proper  to  do  for  the 
sake  of  beginners. 

PROP.  XXV.    B.  VI. 

It  is  very  evident  that  the  demonstration  which  Euclid  had 
given  of  this  proposition  has  been  vitiated  by  some  unskilful  hand: 
for,  after  this  editor  had  demonstrated  that  "as  the  rectilineal 
figure  ABC  is  to  the  rectilineal  KGH,  so  is  the  parallelogram 
BE  to  the  parallelogram  EF;"  nothing  more  should  have  been 
added  but  this,  "  and  the  rectilineal  figure  ABC  is  equal  to  the 
parallelogram  BE:  therefore  the  rectilineal  KGH  is  equal  to  the 
parallelogram  EF,"  viz.  from  prop.  '4,  book  5.  But  betwixt 
these  two  sentences  he  has  inserted  this^  "wherefore,  by  permu- 
tation, as  the  rectilineal  figure  ABC  to  the  parallelogram  BE, 
so  is  the  retilineal  KGH  to  the  parallelogram  EF;"  by  which  it 
is  plain,  he  thought  it  was  not  evident  to  conclude,  that  the 
second  of  four  proportionals  is  equal  to  the  fourth  from  the  equal- 
ity of  the  first  and  third,  which  is  a  thing  demonstrated  in  the 
1 4th  prop,  of  B.  5,  as  to  conclude  that  the  third  is  equal  to  the 
fourth,  from  the  equality  of  the  first  and  second,  which  is  no  where 
demonstrated  in  the  elements  as  we  now  have  them:  but  though 
this  proposition,  viz.  the  third  of  four  proportionals  is  equal  to 
the  fourth,  if  the  first  be  equal  to  the  second,  had  been  given  in 
the  Elements  by  Euclid,  as  very  probably  it  was,  yet  he  would 
not  have  made  use  of  it  in  this  place;  because,  as  was  said,  the 
conclusion  could  have  been  immediately  deduced  without  this 
superfluous  step,  by  permutation:  this  we  have  shown  at  the 
greater  length,  both  because  it  aftbrds  a  certain  proof  of  the  vitia- 
tion of  the  text  of  Euclid;  for  the  very  same  blunder  is  found 
twice  in  the  Greek  text  of  prop.  23,  book  11,  and  twice  in  prop. 
2,  B.  12,  and  in  the  5,  11,  12,  and  l8th  of  that  book;  in  which 
places  of  book  12,  except  the  last  of  them,  it  is  rightly  left  out  in 
the  Oxford  edition  of  Commandine's  translation;  and  also  that 
geometers  may  beware  of  making  use  of  permutation  in  the  like 
cases:  for  the  moderns  not  unfrequently  commit  this  mistake,  and 
among  others  Commandine  himself,  in  his  commentary  on  prop. 
5,  book  3,  p.  6,  b.  of  Pappus  Alexandrinus,  and  in  other  places: 
the  vulgar  notion  of  proportionals  has,  it  seems  pre-occupied 
many  so  much,  that  they  do  not  sufficiently  understand  the  true 
liature  of  them. 

Besides,  though  the  rectilineal  figure  ABC,  to  which  another  is 
to  be  made  similar,  may  be  of  any  kind  whatever;  yet  in  the  de- 
monstration the  Greek  text  has  "  triangle"  instead  of  "rectilineal 
figure,"  which  error  is  corrected  in  the  above  named  Oxford 
edition. 

PROP.  XXVII.    B.  VI. 

The  second  case  of  this  has  otAAcos,  otherwise,  prefixed  to  it,  as 
if  it  was  a  different  demonstration,  which  probably  has  been  done 
by  some  unskilful  librarian.     Dr.  Gregory  has  rightly  left  it  out: 


BOOK  VI.  NOTES.  269 

the  scheme  of  this  second  case  ought  to  be  marked  with  the  same 
letters  of  the  alphabet  which  are  in  the  scheme  of  the  first,  as  is 
now  done. 

PROP.  XXVIII.  and  XXIX.    B.  VI. 

These  two  problems,  to  the  first  of  which  the  27th  prop,  is 
necessary,  are  the  most  general  and  useful  of  all  in  the  Elements, 
and  are  most  frequently  made  use  of  by  the  ancient  geometers 
in  the  solution  of  other  problems;  and  therefore  are  very  igno- 
rantly  left  out  by  Tacquet  and  Dechales  in  their  editions  of  the 
Elements,  who  pretend  that  they  are  scarce  of  any  use.  The 
cases  of  these  problems,  wherein  it  is  required  to  apply  a  rect- 
angle which  shall  be  equal  to  a  given  square,  to  a  given  straight 
line,  either  deficient  or  exceeding  by  a  square;  as  also  to  apply  a 
rectangle  which  shall  be  equal  to  another  given,  to  a  given 
straight  line,  deficient  or  exceeding  by  a  square,  are  very  often 
made  use  of  by  geometers.  And,  on  this  account,  it  is  thought 
proper,  for  the  sake  of  beginners,  to  give  their  constructions  as 
follows: 

1.  To  apply  a  rectangle  which  shall  be  equal  to  a  given  square, 
to  a  given  straight  line,  deficient  by  a  square;  but  the  given  square 
must  not  be  greater  than  that  upon  the  half  of  the  given  line. 

Let  AB  be  the  given  straight  line,  and  let  the  square  upon  the 
given  straight  line  C  be  that  to  which  the  rectangle  to  be  applied 
must  be  equal,  and  this  square,  by  the  determination,  is  not  greater 
than  that  upon  half  of  the  straight  line  AB. 

Bisect  AB  in  D,  and  if  the  square  upon  AD  be  equal  to  the 
square  upon  C,  the  thing  required  is  done:  but  if  it  be  not  equal 
to  it,  AD  must  be  greater  than  C, 
according  to  the  determination;  L 
draw  DE  at  right  angles  to  AB, 
and  make  it  equal  to  C:  produce  A 
ED  to  F,  so  that  EF  be  equal  to 
AD  or  DB,  and  from  the  centre 
E,  at  the  distance  EF,  describe  a 
circle  meeting  AB  in  G,  and  upon 
GB  describe  the  square  GBKH, 
and     complete     the      rectangle  E 

AGHL;  also  join  EG.  And  be- 
cause AB  is  bisected  in  D,  the  rectangle  AG,  GB  together  with 
the  square  of  DG  is  equal  (5.  2.)  to  (the  square  of  DB,  that  is,  of 
EF  or  EG,  that  is,  to)  the  squares  of  ED,  DG:  take  away  the 
square  of  DG  from  each  of  these  equals;  therefore  the  remaining 
rectangle  AG,  GB  is  equal  to  the  square  of  ED,  that  is,  of  C;  but 
the  rectangle  AG,  GB  is  the  rectangle  AH,  because  GH  is  equal 
to  GB;  therefore  the  rectangle  AH  is  equal  to  the  given  squaf^e 
upon  the  straight  line  C.  Wherefore  the  rectangle  AH,  equal  to 
the  given  square  upon  C,  has  been  applied  to  the  given  straight 
line  AB  deficient  by  the  square  GK.     Which  was  to  be  done. 

2.  To  apply  a  rectangle  which  shall  be  equal  to  a  given  square, 
to  a  given  straight  line,  exceeding  by  a  square. 


270 


NOTES. 


BOOK  VI. 


FAD 


B       G 


Let  AB  be  the  given  straii>ht  line,  and  let  the  square  upon  the 
given  straight  line  C  be  that  to  which  the  rectangle  to  be  applied 
must  be  equal. 

Bisect  AB  in  D,  and  draw  BE  at  right  angles  to  it,  so  that  BE 
be  equal  to  C;  and  having  joined  DE,  from  the  centre  D  at  the 
distance  DE  describe  a  circle  meeting  AB  produced  in  G;  upon 
BG  describe  the  square  BGHK,  and  —  ""^^ 

complete  the  rectangle  AGHL.  And 
because  AB  is  bisected  in  D,  and 
produced  to  G,  the  rectangle  AG, 
GB  together  with  the  square  of  DB 
is  equal  (6.  2.)  to  (the  square  of  DG, 
or  DE,  that  is,  to)  the  squares  of  EB, 

BD.    From  each  of  these  equals  take 

the  square  of  DB5  therefore,  the  re-  q 

maining  rectangle  AG,  GB  is  equal 

to  the  square  of  BE,  that  is,  to  the  square  upon  C.  But  the  rect- 
angle AG,  GB  is  the  rectangle  AH,  because  GH  is  equal  to  GB: 
therefore  the  rectangle  AH  is  equal  to  the  square  upon  C. 
Wherefore  the  rectangle  AH,  equal  to  the  given  square  upon  C, 
has  been  applied  to  the  given  straight  line  AB,  exceeding  by  the 
square  GK.     Which  was  to  be  done. 

3.  To  apply  a  rectangle  to  a  given  straight  line  which  shall  be 
equal  to  a  given  rectangle,  and  be  deficient  by  a  square.  But  the 
given  rectangle  must  not  be  greater  than  the  square  upon  the  half 
of  the  given  straight  line. 

Let  AB  be  the  given  straight  line,  and  let  the  given  rectangle 
be  that  which  is  contained  by  the  straight  lines  C,  D  which  is  not 
greater  than  the  square  upon  the  half  of  AB^  it  is  required  to 
apply  to  AB  a  rectangle  equal  to  the  rectangle  C,  D,  deficient  by 
a  square. 

Draw  AE,  BF  at  right  angles  to  AB,  upon  the  same  side  of  it, 
and  make  AE  equal  to  C,  and  BF  t^  D:  join  EF,  and  bisect  it  in 
G;  and. from  the  centre  G,  at  the  distance  GE,  describe  a  circle 
meeting  AE  again  in  H:  join  HF,  and  draw  GK  parallel  to  it,  and 
GL  parallel  to  AE,  meeting  AB  in  L. 

Because  the  angle  EHF  in  a  semicircle  is  equal  to  the  right 
angle  EAB,  AB  and  HF  are  parallels,  and  AH  and  BF  are  paral- 
lels; wherefore  AH  is  equal  to  BF,  and  the  rectangle  EA,  AH 
equal  to  the  rectangle  EA,  BF,  that  is  to  the  rectangle  C,  D:  and 
because  EG,  GF  are  equal  to  one  another,  and  AE,  LG,  BF  pa- 
rallels; therefore  AL  and  LB  are  equal:  also  EK  is  equal  to  KH 
(3.  3.),  and  the  rectangle  C,  D,  from  the  determination,  is  not 
greater  than  the  square  of  AL,  the  half  of  AB;  wherefore  the  rect- 
angle EA,  AH  is  not  greater  than  the  square  of  AL,  that  is  of 
KG:  add  to  each  the  square  of  KE;  therefore  the  square  (6.  2.) 
of  AK  is  not  greater  than  the  squares  of  EK,  KG,  that  is,  than 


BOOK  VI 


NOyES. 


271 


E 


f-.. 


P 


H 


A 


1           --N 

\. 

] 

1^ 

:::-^ 

. 

""^^ 

\ 

/ 

\ 

,M 

I. 

N/ 

^^-—^ 

B 


the  square  of  EG;  and  conse- 
quently the  straight  line  AK  or 
GL  is  not  greater  than  GE. 
Now,  if  GE  be  equal  to  GL,  the 
circle  EHF  touches  AB  in  L, 
and  therefore  the  square  of  AL 
is  (36.  3.)  equal  to  the  rectangle 
EA,  AH,  that  is,  to  the  given 
rectangle  C,  D;  and  that  which 
was  required  is  done:  but  if 
EG,  GL  be  unequal,  EG  must 
be  the  greater:  and  therefore  the 
circle  EHF  cuts  the  straight 
line  AB:   let  it  cut  it  in   the  Q  P     O 

points  M,  N,  and  upon  NB  describe  the  square  NBOP,  and  com- 
plete the  rectangle  ANFQ:  because  LM  is  equal  to  (3.  3.)  LN, 
and  it  has  been  proved  that  AL  is  equal  to  LB;  therefore  AM  is 
equal  to  NB,  and  the  rectangle  AN,  NB  equal  to  the  rectangle 
NA,  AM,  that  is,  to  the  rectangle  (Cor.  36.  3.)  EA,  AH,  or  the 
rectangle  C,  D:  but  the  rectangle  AN,  NB  is  the  rectangle  AP, 
because  PN  is  equal  to  NB:  therefore  the  rectangle  AP  is  equal 
to  the  rectangle  C,  D;  and  the  rectangle  AP  equal  to  the  given 
rectangle  C,  D  has  been  applied  to  the  given  straight  line  AB,  de- 
ficient by  the  square  BP.     Which  was  to  be  done. 

4.  To  apply  a  rectangle  to  a  given  straight  line  that  shall  be 
equal  to  a  given  rectangle,  exceeding  by  a  square. 

Let  AB  be  the  given  straight  line,  and  the  rectangle  C,  D  the 
given  rectangle,  it  is  required  to  apply  a  rectangle  to  AB  equal  to 
CD,  exceeding  by  a  square. 

Draw  AE,  BF  at  right  angles  to  AB,  on  the  contrary  sides  of 


BF  equal  to  D:  join  EF,  and 
centre  G,  at  the  distance  GE,  de- 
H;  join  HF,  and  draw  CL 

E 

/ 


it,  and.  make  AE  equal  to  C,  and 
bisect  it  in  G;  and  from  the 
scribe  a  circle  meeting  AE  again  in 
parallel  to  AE;  let  the  circle 
meet  AB  produced  in  M,  N, 
and  upon  BN  describe  the 
square  NBOP,  and  complete 
the  rectangle  ANPQ;  because 
the  angle  EHF  in  a  semicircle 
is  equal  to  the  right  angle 
EAB,  AB  and  HF  are  paral- 
lels, and  therefore  AH  and  BF 
are  equal,  and  the  rectangle 
EA,  AH  equal  to  the  rectangle 
EA,  BF,  that  is,  to  the  rect- 
angle C,  D:  and  because  ML 

is  equal  to  LN,  and  AL  to  LB,  therefore  MA  is  equal  to  BN,  and 
the  rectangle  AN,  NB  to  MA,  AN,  that  is  (35.  3.)  to  the  rectangle 
EA,  AH,  or  the  rectangle  C,  D:  therefore  the  rectangle  AN,  NB, 
that  is,  AP,  is  equal  to  the  rectangle  C,  D;  and  to  the  given 
straight  line  AB  the  rectangle  AP  has  been  applied  equal  to  the 


272  NOTES.  BOOK  VI. 

given  rectangle  C,  D,  exceeding  by  the  square  BP.     Which  was 
to  be  done. 

Willebrordus  Snellius  was  the  first,  as  far  as  I  know,  who  gave 
these  constructions  of  the  3d  and  4th  problems,  in  his  Apollonius 
Batavus;  and  afterwards  the  learned  Dr.  Halley  gave  them  in  the 
scholium  of  the  18th  prop,  of  the  8th  book  of  Apollonius's  conies 
restored  by  him. 

The  3d  problem  is  otherwise  enunciated  thus:  to  cut  a  given 
straight  line  AB  in  the  point  N,  so  as  to  make  the  rectangle  AN, 
NB  equal  to  a  given  space;  or,  which  is  the  same  thing,  having 
given  AB  the  sum  of  the  sides  of  a  rectangle,  and  the  magnitude 
of  it  being  likewise  given,  to  find  its  sides. 

And  the  fourth  problem  is  the  same  with  this.  To  find  the 
point  N  in  the  given  straight  line  AB  produced,  so  as  to  make 
the  .rectangle  AN,  NB  equal  to  a  given  space:  or,  which  is  the 
same  thing,  having  given  AB  the  difference  of  the  sides  of  a  rect- 
angle, and  the  magnitude  of  it,  to  find  the  sides. 

PROP.  XXXI.     B.  VI. 

In  the  demonstration  of  this,  the  inversion  of  proportionals  is 
twice  neglected,  and  is  now  added,  that  the  conclusion  may  be  le- 
gitimately made  by  help  of  the  24th  prop,  of  B.  5.  as  Clavius  had 
done. 

PROP.  XXXII.     B.  VI. 

The  enunciation  of  the  preceding  26th  prop,  is  not  general 
enough;  because  not  only  two  similar  parallelograms  that  have 
an  angle  common  to  both,  are  about  the  same  diameter;  but  like- 
wise two  similar  parallelograms  that  have  vertically  opposite  an- 
gles, have  their  diameters  in  the  same  straight  line:  but  there 
seems  to  have  been  another,  and  that  a  direct  demonstration  of 
these  cases,  to  which  this  32d  proposition  was  needful:  and  the 
32d  may  be  otherwise  and  something  more  briefly  demonstrated 
as  follows. 

PROP.  XXXII.     B.  VI. 

If  two  triangles  which  have  two  sides  of  the  one.  See. 

Let  GAF,  HFC  be  two  triangles  which  have  two  sides  AG,  GF 
proportional  to  the  two  sides  FH,  HC,  viz.  AG  to  GF,  as  FH  to 
HC;  and  let  AG  be  parallel  to  FH,  and         AG  D 

GF  to  HC|  AF  and  FC  are  in  a  straight 
line. 

Draw  CK  parallel  (31. 1.)  to  FH,  and      E  ^|^^ H 

let  it  meet  GF  produced  in  K:  because 
AG,  KG,  are  each  of  them  parallel  to 
FH,  they  are  parallel   (30.   1.)   to   one 
another,  and  therefore  the  alternate  an- 
gles AGF,  FKC  are  equal:  and  AG  is  ^         ^  ^ 
to  GF,  as  (FH  to  HC,  that  is  34.  i.)  CK  to  KF;  wherefore  the 
triangles  AGF,  CKF  are  equiangular,  (6.  6.)  and  the  angle  AFG 
equal  to  the  angle  CFK:  but  GFK  is  a  straight  line,  therefore 
AF  and  FC  are  in  a  straight  line  (14.  I.). 


Hi 


liOOKXI.  NOTES.  273 

The  26th  prop,  is  demonstrated  from. the  32d,  as  follows: 

If  two  similar  and  similarly  placed  parallelograms  have  an  an- 
gle common  to  both,  or  vertically  opposite  angles^  their  diame- 
ters are  in  the  same  straight  line. 

First,  let  the  parallelograms  ABCD,  AEFG  have  the  angle 
BAD  common  to  both,  and  be  similar  and  similarly  placed^ 
ABCD,  AEFG  are  about  the  same  diameter. 

Produce  EF,  GF,  to  FI,  K,  and  join  FA,  FC^  then  because  the 
parallelograms  ABCD,  AEFG  are  similar,  DA  is  to  AB,  as  GA 
to  AE:  wherefore  the  remainder  DG  is  A  G  D 

(Cor.  19.  5.)  to  the  remainder  EB,  as 
GA  to  AE:  but  DG  is  equal  to  FH, 

EB  to  HC,  and  AE  to  GF:  therefore      E  | :H^: 1  H 

as  FH  to  HC,  so  is  AG  to  GF^  and 

FH,  HC  are  parallel  to  AG,  GF;  and 

the  triangles  AGF,  FHC  are  joined  at 

one  angle  in  the   point  F:    wherefore 

AF,  FC  are  in  the  same  straight  line  B  K  C 

(32.  6.). 

Next,  let  the  parallelograms  KFHC,  GFEA,  -which  are  similar 
and  similarly  placed,  have  their  angles  KFH,  GFE  vertically  op- 
posite; their  diameters  AF,  FC  are  in  the  same  straight  line. 

Because  AG,  GF  are  parallel  to  FH,  HC;  and  that  AG  is  to 
GF,  as  FH  to  HC;  therefore  AF,  FC  are  in  the  same  straight 
line  (32.  6.). 

PROP.  XXXHI.     B.  VI. 

The  words  "because  they  are  at  the  centre,"  are  left  out,  as  the 
addition  of  some  unskilful  hand. 

In  the  Greek,  as  also  in  the  Latin  translation,  the  words  oc  srv^e 
"  any  whatever,"  are  left  out  in  the  demonstration  of  both  parts 
of  the  proposition,  and  are  now  added  as  quite  necessary;  and  in 
the  demonstration  of  the  second  part,  where  the  triangle  BGC  is 
proved  to  be  equal  to  CGK,  the  illative  particle  u^cc  in  the  Greek 
text  ought  to  be  omitted. 

The  second  part  of  the  proposition  is  an  addition  of  Theon's, 
as  he  tells  us  in  his  commentary  on  Ptolemy's  Mey^Aj;  Zwru^i^^ 
p.  50. 

PROP.  B.  C.  D.     B.  VI. 

These  three  propositions  are  added,  because  they  are  frequently 
made  use  of  by  geometers. 

DEF.  IX.  and  XI.     B.  XI. 

The  similitude  of  plane  figures  is  defined  from  the  equality  of 
their  angles,  and  the  proportionality  of  the  sides  about  the  equal 
angles;  for  from  the  proportionality  of  the  sides  only,  or  only 
from  the  equality  of  the  angles,  the  similitude  of  the  figures  does 
not  follow,  except  in  the  case  when  the  figures  are  triangles:  the 
similar  position  of  the  sides  which  contain  the  figures,  to  one  ano- 
ther, depending  partly  upon  each  of  these:  and,  for  the  same  rea- 
son, those  are  similar  solid  figures  which  have  all  their  solid  an- 
gles equal,  each  to  each,  and  are  contained  by  the  same  number 
2  INI 


274  NOTES.  BOOK  XI, 

of  similar  plane  figures:  for  there  are  some  solid  figures  contained 
by  similar  plane  figures,  of  the  same  number,  and  even  of  the 
same  magnitude,  that  are  neither  similar  nor  equal,  as  shall  be 
demonstrated  after  the  notes  on  the  10th  definition;  upon  this  ac- 
count it  M^as  necessary  to  amend  the  definition  of  similar  solid 
figures,  and  to  place  the  definition  of  a  solid  angle  before  it:  and 
from  this  and  the  10th  definition,  it  is  sufficiently  plain  how  much 
the  Elements  have  been  spoiled  by  unskilful  editors. 

DEF.  X.     B.  XI. 

Since  the  meaning  of  the  word  "equal"  is  known  and  esta- 
blished before  it  comes  to  be  used  in  this  definition;  therefore 
the  proposition  which  is  the  10th  definition  of  this  book,  is  a  theo- 
rem, the  truth  or  falsehood  of  which  ought  to  be  demonstrated, 
not  assumed;  so  that  Theon,  or  some  other  editor,  has  ignorantly 
turned  a  theorem  which  ought  to  be  demonstrated  into  this  10th 
definition;  that  figures  are  similar,  ought  to  be  proved  from  the 
definition  of  similar  figures;  that  they  are  equal,  ought  to  be  de- 
monstrated from  the  axiom,  "  Magnitudes  that  wholly  coincide, 
are  equal  to  one  another:"  or  from  prop.  A,  of  book  5,  or  the  9th 
prop,  or  the  14th  of  the  same  book,  from  one  of  which  the  equality 
of  all  kind  of  figures  must  ultimately  be  deduced.  In  the  pre- 
ceding books,  Euclid  has  given  no  definition  of  equal  figures,  and 
it  is  certain  he  did  not  give  this:  for  what  is  called  the  1st  def.  of 
the  3d  book  is  really  a  theorem  in  which  these  circles  are  said  to 
be  equal,  that  have  the  straight  lines  from  their  centres  to  the 
circumferences  equal,  which  is  plain,  from  the  definition  of  a  cir- 
cle; and  therefore  has  by  some  editor  been  improperly  placed 
among  the  definitions.  The  equality  of  figures  ought  not  to  be 
defined,  but  demonstrated:  therefore,  though  it  were  true,  that 
solid  figures  contained  by  the  same  number  of  similar  and  equal 
plane  figures  are  equal  to  one  another,  yet  lie  would  justly  deserve 
to  be  blamed  who  would  make  a  definition  of  this  proposition, 
which  ought  to  be  demonstrated.  But  if  this  proposition  be  not 
true,  must  it  not  be  confessed,  that  geometers  have,  for  these  thir- 
teen hundred  years,  been  mistaken  in  this  elementary  matter? 
And  this  should  teach  us  modesty,  and  to  acknowledge  how  lit- 
tle, through  the  weakness  of  our  minds,  we  are  able  to  prevent 
mistakes,  even  in  the  principles  of  sciences  which  are  justly  reck- 
oned amongst  the  most  certain;  for  that  the  proposition  is  not 
universally  true,  can  be  shown  by  many  examples;  the  following 
is  sufficient. 

Let  there  be  any  plane  rectilineal  figure,  as  the  triangle  ABC, 
and  from  a  point  D  within  it  draw  (12.  11.)  the  straight  line  DE 
at  right  angles  to  the  plane  ABC;  in  DE  take  DE,  DF  equal  to 
one  another,  upon  the  opposite  sides  of  the  plane,  and  let  G  be 
any  point  in  EF;  join  DA,  DB,  DC;  EA,  EB,  EC;  FA,  FB,  FC; 
GA,  GB,  GC;  because  the  straight  line  EDF  is  at  right  angles 
to  the  plane  ABC,  it  makes  right  angles  with  DA,DB,  DC  which 
it  meets  in  that  plane:  and  in  the  triangles  EDB,  FDB,  ED  and 
DB  are  equal  to  FD  and  DB,each  to  each,  and  they  contain  right 


BOOK  XI. 


NOTES. 


275 


B^ 


angles;  therefore  the  base 
EB  is  equal  (4.  1.)  to  the 
base  FB;  in  the  same  man- 
ner EA  is  equal  to  FA,  and 
EC  to  FC:  and  in  the  trian- 
gles EBA,  FBA,  EB,  BA 
are  equal  to  FB,  BA;  and 
the  base  EA,  is  equal  to  the 
base  FA;  wherefore  the  an- 
gle EBA  is  equal  (8.  1.)  to 
the  angle  FBA,  and  the  tri- 
angle EBA  equal  (4.  1.)  to 
the  triangle  FBA,  and  the 
other  angles  equal  to  the 
other  angles;  therefore  these 
triangles  are  similar  (4.  6.  F 

1  def.):  in  the  same  manner 

the  triangle  EBC  is  similar  to  the  triangle  FBC,  and  the  triangle 
EAC  to  FAC;  therefore  there  are  two  solid  figures,  each  of  which 
is  contained  by  six  triangles,  one  of  them  by  three  triangles,  the 
common  vertex  of  which  is  the  point  G,  and  their  bases  the 
straight  lines  AB,  BC,  CA,  and  by  three  other  triangles  the  com- 
mon vertex  of  which  is  the  point  E,  and  their  bases  the  same  lines 
AB,  BC,  CA:  the  other  solid  is  contained  by  the  same  three  tri- 
angles the  common  vertex  of  which  is  G,  and  their  bases  AB,  BC, 
CA;  and  by  three  other  triangles  of  which  the  common  vertex  is 
the  point  F,  and  their  bases  the  same  straight  lines  AB,  BC,  CA: 
now  the  three  triangles  GAB,  GBC,  GCA  are  common  to  both 
solids,  and  the  three  others  EAB,  EBC,  ECA  of  the  first  solid 
have  been  shown  equal  and  similar  to  the  three  others  FAB,  FBC, 
FCA  of  the  other  solid,  each  to  each;  therefore  these  two  solids 
are  contained  by  the  same  number  of  equal  and  similar  planes: 
but  that  they  are  not  equal  is  manifest,  because  the  first  of  them 
is  contained  in  the  other:  therefore  it  is  not  universally  true  that 
solids  are  equal  which  are  contained  by  the  same  number  of  equal 
and  similar  planes. 

Cor.  From  this  it  appears  that  two  unequal  solid  angles  may 
be  contained  by  the  same  number  of  equal  plane  angles. 

For  the  solid  angle  at  B,  which  is  contained  by  the  four  plane 
angles  EBxA.,  EBC,  GBA,  GBC  is  not  equal  to  the  solid  angle  at 
the  same  point  B,  which  is  contained  by  the  four  plane  angles 
FBA,  FBC,  GBA,  GBC;  for  this  last  contains  the  other:  and 
each  of  them  is  contained  by  four  plane  angles  which  are  equal 
to  one  another,  each  to  each,  or  are  the  self  same;  as  has  been 
proved:  and  indeed  there  may  be  innumerable  solid  angles  all 
unequal  to  one  another,  which  are  each  of  them  contained  by 
plane  angles  that  are  equal  to  one  another,  each  to  each:  it  is 
likewise  manifest  that  the  before  mentioned  solids  are  not  similar, 
since  their  solid  angles  are  not  all  equal. 

And  that  there  may  be  innumerable  solid  angles  all  unequal  to 
one  another,  which  are  each  of  them  contained  by  the  same  plane 


276  NOTES.  BOOK  XI. 

angles  disposed  in  the  same  order,  will  be  plain  from  the  three 
following  propositions. 

PROP.  I.     PROBLEM. 

Three  magnitudes,  A,  B,  C  being  given,  to  find  a  fourth  such, 
that  every  three  shall  be  greater  than  the  remaining  one. 

Let  D  be  the  fourth:  therefore  D  must  be  less  than  A,  B,  C  to- 
gether: of  the  three,  A,  B,  C,  let  A  be  that  which  is  not  less  than 
either  of  the  two  B  and  C:  and  first,  let  B  and  C  together  be  not 
less  than  A:  therefore  B,  C,  D  together  are  greater  than  A;  and 
because  A  is  not  less  than  B;  A,  C,  D  together  are  greater  than 
B:  in  the  like  manner.  A,  B,  D  together  are  greater  than  C: 
wherefore,  in  the  case  in  which  B  and  C  together  are  not  less 
than  A,  any  magnitude  D  which  is  less  than  A,  B,  C  together, 
will  answer  the  problem. 

But  if  B  and  C  together  be  less  than  A;  then,  because  it  is  re- 
quired that  B,  C,  D  together  be  greater  than  A,  from  each  of 
these  taking  away  B,  C,  the  remaining  one  D  must  be  greater 
than  the  excess  of  A  above  B  and  C^  take  therefore  any  magni- 
tude D  which  is  less  than  A,  B,  C  together,  but  greater  than  the 
excess  of  A  above  B  and  C:  then  B,  C,  D  together  are  greater 
than  A;  and  because  A  is  greater  than  either  B  or  C,  much  more 
will  A  and  D  together  with  either  of  the  two  B,  C  be  greater  than 
the  others  and  by  the  construction.  A,  B,  C  are  together  greater 
than  D. 

Coil.  If  besides  it  be  required,  that  A  and  B  together  shall  not 
be  less  than  C  and  D  together;  the  excess  of  A  and  B  together 
above  C  must  not  be  less  than  D,  that  is,  D  must  not  be  greater 
than  that  excess. 

PROP.  n.     PROBLEM. 

Four  magnitudes  A,  B,  C,  D  being  given,  of  which  A  and  B 
together  are  not  less  than  C  and  D  together,  and  such  that  any 
three  of  them  whatever  are  greater  than  the  fourth;  it  is  required 
to  find  a  fifth  magnitude  E  such,  that  any  two  of  the  three  A,  B, 
E  shall  be  greater  than  the  third,  and  also  that  any  two  of  the 
three  C,  D,  E  shall  be  greater  than  the  third.  Let  A  be  not  less 
than  B,  and  C  not  less  than  D. 

First,  let  the  excess  of  C  above  D  be  not  less  than  the  excess 
of  A  above  B:  it  is  plain  that  a  magnitude  E  can  be  taken  which 
is  less  than  the  sum  of  C  and  D,  but  greater  than  the  excess  of  C 
above  D;  let  it  be  taken;  then  E  is  greater  likewise  than  the  ex- 
cess of  A  above  B:  wherefore  E  and  B  together  are  greater  than 
A;  and  A  is  not  less  than  B;  therefore  A  and  E  together  are 
greater  than  B:  and,  by  the  hypothesis,  A  and  B  together  are  not 
less  than  C  and  D  together,  and  C  and  D  together  are  greater 
than  E;  therefore  likewise  A  and  B  are  greater  than  E. 

But  let  the  excess  of  A  above  B  be  greater  than  the  excess  of 
C  above  D;  and  because,  by  the  hypothesis,  the  three  B,  C,  D  are 
together  greater  than  the  fourth  A:  C  and  D  together  are  greater 
than  the  excess  of  A  above  B:  therefore  a  magnitude  may  be 
taken  which  is  less  than  C  and  D  together,  but  greater  than  the 


1! 


BOOK  XI.  NOTES.  277 

excess  of  A  above  B.  Let  this  magnitude  be  E;  and  because  E 
is  greater  than  the  excess  of  A  above  B,  B  together  with  E  is 
greater  than  A:  and,  as  in  the  preceding  case,  it  may  be  shown 
that  A  together  with  E  is  greater  than  B,  and  that  A  together 
with  B  is  greater  than  E:  therefore,  in  each  of  the  cases,  it  has 
been  shown  that  any  two  of  the  three  A,  B,  E  are  greater  than 
the  third'. 

And  because  in  each  of  the  cases  E  is  greater  than  the  excess 
of  C  above  D,  E  together  with  D  is  greater  than  C;  and,  by  the 
hypothesis,  C  is  not  less  than  D;  therefore  E  together  with  C  is 
greater  than  D;  and,  by  the  construction,  C  and  D  together  are 
greater  than  E:  therefore  any  two  of  the  three,  C,  D,  E  are  greater 
than  the  third. 

PROP.  III.     THEOREM. 

There  may  be  innumerable  solid  angles  all  unequal  to  one  ano- 
ther, each  of  which  is  contained  by  the  same  four  plane  angles, 
placed  in  the  same  order. 

Take  three  plane  angles  A,  B,  C,  of  which  A  is  not  less  than 
either  of  the  other  two,  and  such,  that  A  and  B  together  are  less 
than  two  right  angles:  and  by  problem  1,  and  its  corollary,  find  a 
fourth  angle  D  such,  that  any  three  whatever  of  the  angles  A,  B, 
C,  D  be  greater  than  the  remaining  angle,  and  such,  that  A  and 
B  together  be  not  less  than  C  and  D  together:  and  by  problem  2, 
find  a  fifth  angle  E  such,  that  any  two  of  the  angles  A,  B,  E  be 
greater  than  the  third,  and  also  that  any  two  of  the  angles  C,  D, 

A  E  C  F 


E,  be  greater  than  the  third:  and  because  A  and  B  together  are 
less  than  two  right  angles,  the  double  of  A  and  B  together  is  less 
than  four  right  angles:  but  A  and  B  together  are  greater  than 
the  angle  E;  wherefore  the  double  of  A,  B  together  is  greater 
than  the  three  angles  A,  B,  E  together,  which  three  are  conse- 
quently less  than  four  right  angles^  and  every  two  of  the  same 
angles  A,  B,  E  are  greater  than  the  third;  therefore,  by  prop.  23. 
11,  a  solid  angle  may  be  made  contained  by  three  plane  angles 
equal  to  the  angles  A,  B,  E,  each  to  each.  Let  this  be  the  angle 
F  contained  by  the  three  plane  angles  GFH,  HFK,  GFK  which 
are  equal  to  the  angles  A,  B,  E,  each  to  each:  and  because  the 
angles  C,  D  together  are  not  greater  than  the  angles  A,  B  toge- 
ther, therefore  the  angles  C,  D,  E  are  not  greater  than  the  angles 
A,  B,  E:  but  these  last  three  are  less  than  four  right  angles,  as 
has  been  demonstrated:  wherefore  also  the  angles  C,  D,  E  are  to- 


278 


NOTES. 


BOOK  XI. 


gether  less  than  four  right  angles,  and  every  two  of  them  are 
greater  than  the  third;  therefore  a  solid  angle  may  be  made 
which  shall  be  contained  by  three  plane  angles  equal  to  the  an- 
gles C,  D,  E,  each  to  each  (23.  11.):  and  by  prop.  26.  11,  at  the 

A  E  C  F 


H 

point  F  in  the  straight  line  FG  a  solid  angle  may  be  made  equal 
to  that  which  is  contained  by  the  three  plane  angles  that  are  equal 
to  the  angles  C,  D,  E:  let  this  be  made,  and  let  the  angle  GFK, 
which  is  equal  to  E,  be  one  of  the  three;  and  let  KFL,  GFL  be 
the  other  two  which  are  equal  to  the  angles  C,  D,  each  to  each. 
Thus  there  is  a  solid  angle  constituted  at  the  point  F,  contained 
by  the  four  plane  angles  GFH,  HFK,  KFL,  GFL  which  are  equal 
to  the  angles  A,  B,  C,  D,  each  to  each. 

Again,  find  another  angle  M  such,  that  every  two  of  the  three 
angles  A,  B,  M  be  greater  than  the  third,  and  also  every  two  of 
the  three  C,  D,  M  be  greater  than  the  third:  and,  as  in  the  pre- 
ceding part,  it  may  be  demonstrated  that  the  three  A,  B,  M  are 
less  than  four  right  angles,  as  also  N 

that  the  three  C,  D,  M  are  less 
than  four  right  angles.  Make 
therefore  (23.  11.)  a  solid  angle  at 
N  contained  by  the  three  plane  an- 
gles ONP,  PNQ,  ONQ,  which  are 
equal  to  A,  B,  M,  each  to  each: 
and  by  prop.  26.  11,  make  at  the  O 
point  N  in  the  straight  line  ON  a 
solid  angle  contained  by  three  plane  angles  of  which  one  is  the 
angle  ONQ  equal  to  M,  and  the  other  two  are  the  angles  QNR, 
ONR,  which  are  equal  to  the  angles  C,  D,  each  to  each.  Thus, 
at  the  point  N,  there  is  a  solid  angle  contained  by  the  four  plane 
angles  ONP,  PNQ,  QNR,  ONR  which  are  equal  to  the  angles  A, 
B,  C,  D,  each  to  each.  And  that  the  two  solid  angles  at  the 
points  F,  N,  each  of  which  is  contained  by  the  above  named  four 
plane  angles,  are  not  equal  to  one  another,  or  that  they  cannot  co- 
incide, will  be  plain  by  considering  that  the  angles  GFK,  ONQ, 
that  is,  the  angles  E,  M,  are  unequal  by  the  construction;  and 
therefore  the  straight  lines  GF,  FK  cannot  coincide  with  ON, 
NQ,  nor  consequently  can  the  solid  angles,  vrhich  therefore  are 
unequal. 

And  because  from  the  four  plane  angles  A,  B,  C,  D,  there  can 
be  found  innumerable  other  angles  that  will  serve  the  same  pur- 


BOOK  XI.  NOTES.  279 

pose  with  the  angles  E  and  M;  it  is  plain  that  innumerable  other 
solid  angles  may  be  constituted  which  are  each  contained  by  the 
same  four  plane  angles,  and  all  of  them  unequal  to  one  another. 
Q.  E.  D. 

And  from  this  it  appears  that  Clavius  and  other  authors  are 
mistaken,  who  assert  that  those  solid  angles  are  equal  which  are 
contained  by  the  same  number  of  plane  angles  that  are  equal  to 
one  another,  each  to  each.  Also  it  is  plain  that  the  26th  prop,  of 
book  11,  is  by  no  means  sufficiently  demonstrated,  because  the 
equality  of  two  solid  angles,  whereof  each  is  contained  by  three 
plane  angles  which  are  equal  to  one  another,  each  to  each,  is  only 
assumed,  and  not  demonstrated. 

PROP.  I.     B.  XL 

The  words  at  the  end  of  this,  "  for  a  straight  line  cannot  meet 
a  straight  .line  in  more  than  one  point,"  are  left  out,  as  an  addi- 
tion by  some  unskilful  hand 5  for  this  is  to  be  demonstrated,  not 
assumed. 

Mr.  Thomas  Simson,  in  his  notes  at  the  end  of  the  2d  edition 
of  his  Elements  of  Geometry,  p.  262,  after  repeating  the  words  of 
this  note,  adds,  "  Now,  can  it  possibly  show  any  want  of  skill  in 
an  editor  (he  means  Euclid  or  Theon)  to  refer  to  an  axiom  which 
Euclid  himself  hath  laid  down,  book  1 ,  No.  1 4,  (he  means  Barrow's 
Euclid,  for  it  is  the  10th  in  the  Greek),  and  not  to  have  demon- 
strated, what  no  man  can  demonstrate?"  But  all  that  in  this  case 
can  follow  from  that  axiom  is,  that,  if  two  straight  lines  could 
meet  each  other  in  two  points,  the  parts  of  them  betwixt  these 
points  must  coincide,  and  so  they  would  have  a  segment  betwixt 
these  points  common  to  both.  Now,  as  it  has  not  been  shown  in 
Euclid,  that  they  cannot  have  a  common  segment,  this  does  not 
prove  that  they  cannot  meet  in  two  points,  from  which  their  not 
having  a  common  segment  is  deduced  in  the  Greek  edition:  but, 
on  the  contrary,  because  they  cannot  have  a  common  segment,  as 
is  shown  in  cor.  of  11th  prop,  book  1,  of  4to  edition,  it  follows 
plainly  that  they  cannot  meet  in  two  points,  which  the  remarker 
says  no  man  can  demonstrate. 

Mr.  Simson,  in  the  same  notes,  p.  265,  justly  observes,  that  in 
the  corollary  of  prop.  11,  book  1,  4to  edition,  the  straight  lines 
AB,  BD,  BC  are  supposed  to  be  all  in  the  same  plane,  v/hich 
cannot  be  assumed  in  1st  prop,  book  1 1.  This,  soon  after  the  4to 
edition  was  published,  I  observed,  and  corrected  as  it  is  now  in 
this  edition:  he  is  mistaken  in  thinking  the  10th  axiom  he  men- 
tions here  to  be  Euclid's^  it  is  none  of  Euclid's,  but  is  the  10th  in 
Dr.  Barrow's  edition,  who  had  it  from  Herigon's  Cursus,  vol.  1, 
and  in  place  of  it  the  corollary  of  10th  prop,  book  1,  was  added. 

PROP.  II.    B.  XI. 

This  proposition  seems  to  have  been  changed  and  vitiated  by 
some  editor:  for  all  the  figures  defined  in  the  1st  book  of  the  Ele- 
ments, and  among  them  triangles,  are,  by  the  hypothesis,  plane 
figures;  that  is,  such  as  arc  described  in  a  plane;  wherefore  the 
second  part  of  the  enunciation  needs  no  demonstration.     Besides, 


280  NOTES. 


BOOK  XI. 


a  convex  superficies  may  be  terminated  by  three  straight  lines 
meeting  one  another;  the  thing  that  should  have  been  demon- 
strated is,  that  two  or  three  straight  lines,  that  meet  one  anothePj 
are  in  one  plane.  And  as  this  is  not  sufficiently  done,  the  enuncia- 
tion and  demonstration  are  changed  into  those  now  put  into  the  text. 

PROP.  III.     B.  XI. 

In  this  proposition  the  following  words  near  to  the  end  of  it  are 
left  out,  viz.  "  therefore  DEB,  DFB  are  not  straight  lines;  in  the 
like  manner  it  may  be  demonstrated  that  there  can  be  no  other 
straight  line  between  the  points  D,  B;"  because  from  this  that  two 
lines  include  a  space,  it  only  follows  that  one  of  them  is  not  a 
straight  line:  and  the  force  of  the  argument  lies  in  this,  viz.  if  the 
common  section  of  the  planes  be  not  a  straight  line,  then  two 
straight  lines  could  include  a  space,  which  is  dbsurd;  therefore 
the  common  section  is  a  straight  line. 

PROP.  IV.     B.  XI. 

The  words  "  and  the  triangle  AED"  to  the  triangle  BEC"  are 
omitted,  because  the  whole  conclusion  of  the  4th  prop,  book  1,  has 
been  so  often  repeated  in  the  preceding  books,  it  was  needless  to  re- 
peat it  here. 

PROP.  V.    B.  XI. 

In  this,  near  to  the  end,  eTriTre^ej,  ought  to  be  left  out  in  the  Greek 
text:  and  the  word  "  plane"  is  rightly  left  out  in  the  Oxford  edi- 
tion of  Commandine's  translation. 

PROP.  VII.     B.  XI. 

This  proposition  has  been  put  into  this  book  by  some  unskilful 
editor,  as  is  evident  from  this,  that  straight  lines  which  are  drawn 
from  one  point  to  another  in  a  plane,  are  in  the  preceding  books, 
supposed  to  be  in  that  plane:  and  if  they  were  not,  some  demon- 
strations in  which  one  straight  line  is  supposed  to  meet  another 
would  not  be  conclusive,  because  these  lines  would  not  meet  one 
another:  for  instance,  in  prop.  30,  book  1,  the  straight  line  GK 
would  not  meet  EF,  if  GK  were  not  in  the  plane  in  which  are  the 
parallels  AB,  CD,  and  in  which,  by  hypothesis,  the  straight  line 
EF  is;  because,  this  7th  proposition  is  demonstrated  by  the  prece- 
ceding  3d,  in  which  the  very  thing  which  is  proposed  to  be  demon- 
strated in  the  7th,  is  twice  assumed,  viz.  that  the  straight  line 
drawn  from  one  point  lo  another  in  a  plane,  is  in  that  plane;  and 
the  same  thing  is  assumed  in  the  preceding  6th  prop,  in  which  the 
straight  line  which  joins  the  points  B,  D  that  are  in  the  plane  to 
which  AB  and  CD  are  at  right  angles,  is  supposed  to  be  in  that 
plane:  and  the  7th,  of  which  another  demonstration  is  given,  is 
kept  in  the  book  merely  to  preserve  the  number  of  the  proposi- 
tions; for  it  is  evident  from  the  7th  and  35th  defiinitions  of  the  1st 
book,  though  it  had  not  been  in  the  Elements. 

PFiOP.  VIII.     B.  XI. 

In  the  Greek,  and  in  Commandine's  and  Dr.  Gregory's  transla- 
tions, near  to  the  end  of  this  proposition,  are  the  following  words; 


^r 


BOOK  XI.  NOTES.  281 

"  but  DC  is  in  the  plane  through  BA,  AD,"  instead  of  which,  in 
the  Oxford  edition  of  Cornmandine's  translation,  is  rightly  put 
"but  UC  is  in  the  plane  through  BD,  DA:"  but  all  the  editions 
have  the  following-  words,  viz.  "  because  AB,  BD  are  in  the  plane 
through  BD,  DA,  and  DC  is  in  the  plane  in  which  are  AB,  BD," 
which  are  manifestly  corrupted,  or  have  been  added  to  the  text; 
for  there  was  not  the  least  necessity  to  go  so  far  about  to  show 
that  DC  is  in  the  same  plane  in  which  are  BD,  DA,  because  it  im- 
mediately follows  from  prop.  7,  preceding,  that  BD,  DA  are  in  the 
plane  in  which  are  the  parallels  AB,  CD:  therefore,  instead  of 
these  words,  there  ought  only  to  be  "  because  all  three  are  in  the 
plane  in  which  are  the  parallels  AB,  CD." 

PROP.  XV.     B.  XI. 

After  the  words  "  and  because  BA  is  parallel  to  GH,"  the  fol- 
lowing are  added,  "  for  each  of  them  is  parallel  to  DE,  and  are  not 
both  in  the  same  plane  with  it,"  as  being  manifestly  forgotten  to 
be  put  into  the  text. 

PROP.  XVI.    B.  XI. 

In  this  near  to  the  end,  instead  of  the  words  "  but  straight  lines 
which  meet  neither  way,"  ought  to  be  read,  "  but  straight  lines  in 
the  same  plane  which  produced  meet  neither  way;"  because, 
though  in  citing  this  definition  in  prop.  27,  book  1,  it  was  not  ne- 
cessary to  mention  the  words, "in  the  same  plane,"  all  the  straight 
lines  in  the  books  preceding  this  being  in  the  same  plane,  yet  here 
it  was  quite  necessary. 

PROP.  XX.     B.  XI. 

In  this,  near  the  beginning,  are  the  words,  "But  if  not  let  BAC 
be  the  greater;"  but  the  angle  BAC  may  happen  to  be  equal  to 
one  of  the  other  two:  wherefore  this  place  should  be  read  thus, 
"  But  if  not,  let  the  angle  BAC  be  not  less  than  either  of  the  other 
two,  but  greater  than  DAB." 

At  the  end  of  this  proposition  it  is  said,  "in  the  same  manner 
it  may  be  demonstrated,"  though  there  is  no  need  of  any  demon- 
stration; because  the  angle  BAC  being  not  less  than  either  of  the 
other  two,  it  is  evident  that  BAC  together  with  one  of  them  is 
greater  than  the  other. 

PROP.  XXII.     B.  XI. 

And  likewise  in  this,  near  the  beginning,  it  is  said,  "  but  if  not, 
let  the  angles  at  B,  E,  H  be  unequal,  and  let  the  angle  at  B  be 
greater  than  either  of  those  at  Eli:"  which  words  manifestly 
show  this  place  to  be  vitiated,  because  the  angle  at  B  may  be 
equal  to  one  of  the  other  two.  They  ought  therefore  to  be  read 
thus,  "But  if  not,  let  the  angles  at  B,  E,  H  be  unequal,  and  let 
the  angle  at  B  be  not  less  than  either  of  the  other  two  at  E,  H: 
therefore  the  straight  line  AC  is  not  less  than  either  of  the  two 
DF,  GK." 

PROP.  XXIII.    B.  XI. 

The  demonstration  of  this  is  made  something  shorter,  by  not 
repeating  in  the  third  case  the  things  which  were  demonstrated 

2  N 


282  NOTES.  BOOK    XI. 

in  the  first;  and  by  making  use  of  the  construction  which  Cam- 
panus  has  given;  but  he  does  not  demonstrate  the  second  and 
third  cases:  the  construction  and  demonstration  of  the  third  case 
are  made  a  little  more  simple  than  in  the  Greek  text. 

PROP.  XXIV.    B.  XI. 

The  word  "  similar"  is  added  to  the  enunciation  of  this  propo- 
sition, because  the  planes  containing  the  solids  which  are  to  be 
demonstrated  to  be  equal  to  one  another,  in  the  25th  proposition, 
ought  to  be  similar  and  equal,  that  the  equality  of  the  solids  may 
be  inferred  from  prop.  C,  of  this  book;  and,  in  the  Oxford  edition 
of  Commandine's  translation,  a  corollary  is  added  to  prop.  24,  to 
show  that  the  parallelograms  mentioned  in  this  proposition  are 
similar,  that  the  equality  of  the  solids  in  prop.  25,  may  be  de- 
duced from  the  10th  def.  of  book  11. 

PROP.  XXV.  and  XXVI.    B.  XI. 

In  the  25th  prop,  solid  figures,  which  are  contained  by  the 
same  number  of  similar  and  equal  plane  figures,  are  supposed  to 
be  equal  to  one  another.  And  it  seems  that  Theon,  or  some 
other  editor,  that  he  might  save  himself  the  trouble  of  demon- 
strating the  solid  figures  mentioned  in  this  proposition  to  be 
equal  t*o  one  another,  has  inserted  the  10th  def.  of  this  book,  to 
serve  instead  of  a  demonstration;  which  was  very  ignorantly 
done. 

Likewise  in  the  26th  prop,  two  solid  angles  are  supposed  to  be 
equal:  if  each  of  them  be  contained  by  three  plane  angles  which 
are  equal  to  one  another,  each  to  each.     And  it  is  strange  enough, 
that  none  of  the  commentators  on  Euclid  have,  as  far  as  I  know, 
perceived    that  something  is  wanting   in  the  demonstrations  of 
these  two  propositions.     Clavius,  indeed,  in  a  note  upon  the  1 1th 
def.  of  this  book,  affirms,  that  it  is  evident  that  those  solid  angles 
are  equal  which   are  contained  by   the  same  number  of  plane 
angles,  equal  to  one  another,  each  to  each,  because  they  will  co- 
incide, if  they  be  conceived  to  be  placed  within  one  another;  but 
this  is  said  without  any  proof,  nor  is  it  always  true,  except  when 
the  solid  angles  are  contained  by  three  plane  angles  only,  which 
are  equal  to  one  another,  each  to  each:  and  in  this  case  the  pro- 
position is  the  same  with  this,  that  two  spherical  triangles  that 
are  equilateral  to  one  another,  are  also  equiangular  to  one  another, 
and  can  coincide;  which  ought  not  to  be  granted  without  a  de- 
monstration.    Euclid  does  not  assume  this  in  the  case  of  recti- 
lineal  triangles,  but  demonstrates,  in  prop.  8,  book   1,  that  tri- 
angles which  are  equilateral  to  one  another  are  also  equiangular 
to   one  another;  and  from  this  their  total  equality  appears  by 
prop.  4,  book  1.     And  Menelaus,  in  the  fourth  prop,  of  his  1st 
book  of  spherics,  explicity  demonstrates,  that  spherical  triangles 
which  are  mutually  equilateral,  are  also  equiangular  to  one  an- 
other; from  which  it  is  easy  to  show  that  they  must  coincide, 
providing  they  have  their  sides  disposed  in  the  same  order  and 
situation. 

To  supply  these  defects,  it  was  necessary  to  add  the  three  pro- 


BOOK  XI.  NOTES.  28S 

positions  marked  A,  B,  C  to  this  book.  For  the  25th,  26th  and 
28th  propositions  of  it,  and  consequently  eight  others,  viz.  the 
27th,  31st,  32d,  33d,  34th,  36th,  37th,  and  40th  of  the  same,  which 
depend  upon  them,  have  hitherto  stood  upon  an  infirm  founda- 
tion; as  also,  the  8th,  12th,  cor.  of  17th  and  18th  of  12th  book, 
which  depend  upon  the  9th  definition.  For  it  has  been  shown  in 
the  notes  on  def.  10th  of  this  book,  that  solid  figures  which  arc 
contained  by  the  same  number  of  similar  and  equal  plane  figures, 
as  also  solid  angles  that  are  contained  by  the  same  number  of 
equal  plane  angles,  are  not  always  equal  to  one  another. 

It  is  to  be  observed  that  Tacquet,  in  his  Euclid,  defines  equal 
solid  angles  to  be  such,  "  as  being  put  within  one  another  do  coin- 
cide;" but  this  is  an  axiom,  not  a  definition;  for  it  is  true  of  all 
magnitudes  whatever,  He  made  this  useless  definition,  that  by  it 
he  might  demonstrate  the  36th  prop,  of  this  book,  without  the 
help  of  the  35th  of  the  same:  concerning  which  demonstration, 
see  the  note  upon  prop.  36. 

PROP.  XXVIII.    B.  XI. 

In  this  it  ought  to  have  been  demonstrated,  not  assumed,  that 
the  diagonals  are  in  one  plane.     Clavius  has  supplied  this  defect. 

PROP.  XXIX.    B.  XI. 

There  are  three  cases  of  this  proposition:  the  first  is,  when  the 
two  parallelograms  opposite  to  the  base  AB  have  a  side  common 
to  both;  the  second  is,  when  these  parallelograms  are  separated 
from  one  another;  and  the  third,  when  there  is  a  part  of  them 
common  to  both;  and  to  this  last  only,  the  demonstration  that  has 
hitherto  been  in  the  Elements  does  agree.  The  first  case  is  im- 
mediately deduced  from  the  preceding  28th  prop,  which  seems 
for  this  purpose  to  have  been  premised  to  this  29th,  for  it  is  ne- 
cessary to  none  but  to  it,  and  to  the  40th  of  this  book,  as  we  now 
have  it,  to  which  last  it  would,  without  doubt,  have  been  pre- 
mised, if  Euclid  had  not  made  use  of  it  in  the  29th;  but  some  un- 
skilful editor  has  taken  it  away  fromx  the  Elements,  and  has  muti- 
lated Euclid's  demonstration  of  the  other  two  cases,  which  is  now 
restored,  and  serves  for  both  at  once. 

PROP.  XXX.    B.  XI. 

In  the  demonstration  of  this,  the  opposite  planes  of  the  solid 
CP,  in  the  figure  in  this  edition,  that  is  of  the  solid  CO  in  Com- 
mandine's  figure,  are  not  proved  to  be  parallel;  which  it  is  proper 
to  do  for  the  sake  of  learners. 

PROP.  XXXI.    B.  XI. 

There  are  two  cases  of  this  proposition:  the  first  is,  when  the 
insisting  straight  lines  are  right  angles  to  the  bases;  the  other, 
when  they  are  not:  the  first  case  divided  again  into  two  others, 
one  of  which  is,  when  the  bases  are  equiangular  parallelograms; 
the  other,  when  they  are  not  equiangular:  the  Greek  editor  makes 
no  mention  of  the  first  of  these  two  last  cases,  but  has  inserted  the 
demonstration  of  it  as  a  part  of  that  of  the  other;  and  therefore 


284  -  NOTES.  BOOK  XI. 

should  have  taken  notice  of  it  in  a  corollary;  but  we  thought  it 
better  to  give  these  two  cases  separately;  the  demonstration  also 
is  made  something  shorter  by  following  the  way  Euclid  has  made 
use  of  in  prop.  14,  book  6.  Besides,  in  the  demonstration  of  the 
case  in  which  the  insisting  straight  lines  are  not  at  right  angles 
to  the  bases,  the  editor  does  not  prove  that  the  solids  described 
in  the  construction  are  parallelopipeds,  which  it  is  not  to  be 
thought  that  Euclid  neglected:  also  the  words  "  of  which  the  in- 
sisting straight  lines  are  not  in  the  same  straight  lines,"  have  been 
added  by  some  unskilful  hand;  for  they  may  be  in  the  same 
straight  lines. 

PROP.  XXXII.    B.  XL 

The  editor  has  forgot  to  order  the  parallelograin  FH  to  be  ap- 
plied in  the  angle  FGH  equal  to  the  angle  LCG,  which  is  neces- 
sary.    Clavius  has  supplied  this. 

Also,  in  the  construction,  it  is  required  to  complete  the  solid  of 
which  the  base  is  FH,  and  altitude  the  same  with  that  of  the 
solid  CD:  but  this  does  not  determine  the  solid  to  be  completed, 
since  there  may  be  innumerable  solids  upon  the  same  base,  and  of 
the  same  altitude:  it  ought  therefore  to  be  said,  "  complete  the 
solid  of  which  the  l)ase  is  FH,  and  one  of  its  insisting  straight 
lines  is  FD;"  the  same  correction  must  be  made  in  the  following 
proposition,  33. 

PROP.  D.    B.  XI. 

It  is  very  probable  that  Euclid  gave  this  proposition  a  place  in 
the  Elements,  since  he  gave  the  like  proposition  concerning  equi- 
angular parallelograms  in  the  23d,  B.  6. 

PROP.  XXXIV.    B.  XI. 

In  this  the  words  av  ai  ecpsi-raToti  ax.  sttriv  ts-i  rav  uvrav  evOeiuv,  "  of 
which  the  insisting  straight  lines  are  not  in  the  same  straight 
lines,'*  are  thrice  repeated;  but  these  words  ought  either  to  be 
left  out,  as  they  are  by  Clavius,  or,  in  place  of  them,  ought  to  be 
put,  "  whether  the  insisting  straight  lines  be,  or  be  not,  in  the 
same  straight  lines:"  for  the  other  case  is  without  any  reason  ex- 
cluded; also  the  words  av  rcc  v-^ti,  "of  which  the  altitudes,"  are 
twice  put  for  m  xt'  tcpzcrao-ctiy  "  of  which  the  insisting  straight 
lines;"  which  is  a  plain  mistake:  for  the  altitude  is  always  at  right 
angles  to  the  base. 

PROP.  XXXV.    B.  XI. 

The  angles  ABH,  DEM  are  demonstrated  to  be  right  angles 
in  a  shorter  way  than  in  the  Greek;  and  in  the  same  way  ACH, 
DFM  may  be  demonstrated  to  be  right  angles:  also  the  repetition 
of  the  same  demonstration,  which  begins  with  "  in  the  same  man- 
ner," is  left  out,  as  it  was  probably  added  to  the  text  by  some 
editor;  for  the  words,  "  in  like  manner  we  may  demonstrate,"  are 
not  inserted  except  when  the  demonstration  is  not  given,  or  when 
it  is  something  different  from  the  other,  if  it  be  given,  as  in  prop. 
26,  of  this  book. — Campanus  has  not  this  repetition. 

We  have  given  another  demonstration  of  the  corollary,  besides 


BOOK  XI. 


NOTES.  285 


the  one  in  the  original,  by  help  of  which  the  following  36th  prop, 
may  be  demonstrated  without  the  35th. 

PRdP.  XXXVI.    B.  XI. 

Tacquet  in  his  Euclid  demonstrates  this  proposition  without 
the  help  of  the  35th^  but  it  is  plain,  that  the  solids  mentioned  in 
the  Greek  text  in  the  enunciation  of  the  proposition  as  equiangu- 
lar, are  such  that  their  solid  angles  are  contained  by  three  plane 
angles  equal  to  one  another,  each  to  each;  as  is  evident  from  the 
construction.  Now  Tacquet  does  not  demonstrate,  but  assumes 
these  solid  angles  to  be  equal  to  one  another;  for  he  supposes  the 
solids  to  be  already  made,  and  does  not  give  the  construction  by 
which  they  are  made:  but,  by  the  second  demonstration  of  the 
preceding  corollary,  his  demonstration  is  rendered  legitimate 
likewise  in  the  case  where  the  solids  are  constructed  as  in  the 
text. 

PROP.  XXXVII.    B.  XI. 

In  this  it  is  assumed,  that  the  ratios  which  are  triplicate  of 
those  ratios  which  are  the  same  with  one  another,  are  likewise 
the  same  with  one  another;  and  that  those  ratios  are  the  same 
with  one  another,  of  v/hich  the  triplicate  ratios  are  the  same  with 
one  another;  but  this  ought  not  to  be  granted  without  a  demon- 
stration; nor  did  Euclid  assume  the  first  and  easiest  of  these  two 
propositions,  but  demonstrated  it  in  the  case  of  duplicate  ratios, 
in  the  22d  prop,  book  6.  On  this  account,  another  demonstration 
is  given  of  this  proposition  like  to  that  which  Euclid  gives  in 
prop.  22,  book  6,  as  Clavius  has  done. 

PROP.  XXXVIII.   B.  XI. 

When  it  is  required  to  draw  a  perpendicular  from  a  point  in 
one  plane,  which  is  at  right  angles  to  another  plane,  unto  this  last 
plane,  it  is  done  by  drawing  a  perpendicular  from  the  point  to- 
the  common  section  of  the  planes;  for  this  perpendicular  will  be 
perpendicular  to  the  plane  by  def.  4,  of  this  book:  and  it  would 
be  foolish  in  this  case  to  do  it  by  the  11  th  prop,  of  the  same:  but 
Euclid  (17,  12,  in  other  editions),  Apollonius,  and  other  geometers, 
when  they  have  occasion  for  this  problem,  direct  a  perpendicular 
to  be  drawn  from  the  point  to  the  plane,  and  conclude  that  it  will 
fall  upon  the  common  section  of  the  planes,  because  this  is  the 
very  same  thing  as  if  they  had  made  use  of  the  construction  above 
mentioned,  and  then  concluded  that  the  straight  line  must  be  per- 
pendicular to  the  plane;  but  is  expressed  in  fewer  words.  Some 
editor,  not  perceiving  this,  thought  it  was  necessary  to  add  this 
proposition,  which  can  never  be  of  any  use  to  the  1 1th  book,  and 
its  being  near  to  the  end  among  propositions  with  which  it  has 
no  connexion,  is  a  mark  of  its  having  been  added  to  the  text. 

PROP.  XXXIX.    B.  XI. 

In  this  it  is  supposed,  that  the  straight  lines  which  bisect  the 
sides  of  the  opposite  planes,  are  in  one  plane,  which  ought  to  have 
been  demonstrated;  as  is  noAv  done. 


286  ^  NOTES.  BOOK  XII. 

BOOK  XII. 

The  learned  Mr.  Moore,  professor  of  Greek  in  the  University 
of  Glasgow,  observed  to  me,  that  it  plainly  appears  from  Ar- 
chimedes's  epistle  to  Dositheus,  prefixed  to  his  books  of  the 
Sphere  and  Cylinder,  which  epistle  he  has  restored  from  ancient 
manuscripts,  that  Eudoxus  was  the  author  of  the  chief  proposi- 
tions in  this  12th  book. 

PROP.  II.    B.  XII. 

At  the  beginning  of  this  it  is  said,  "  if  it  be  not  so,  the  square 
of  BD  shall  be  to  the  square  of  FH,  as  the  circle  ABCD  is  to 
some  space  either  less  than  the  circle  EFGH,  or  greater  than 
it."  And  the  like  is  to  be  found  near  to  the  end  of  this  proposi- 
tion, as  also  in  prop.  5,  11,  12,  18,  of  this  book:  concerning  which, 
it  is  to  be  observed,  that,  in  the  demonstration  of  theorems,  it  is 
sufficient,  in  this  and  the  like  cases,  that  a  thing  made  use  of  in 
the  reasoning  can  possibly  exist,  providing  this  be  evident,  though 
it  cannot  be  exhibited  or  found  by  a  geometrical  construction:  so, 
in  this  place  it  is  assumed,  that  there  may  be  a  fourth  proportional 
to  these  three  magnitudes,  viz.  the  squares  of  BD,  FH,  and  the 
circle  ABCD;  because  it  is  evident  that  there  is  some  square  equal 
to  the  circle  ABCD  though  it  cannot  be  found  geometrically:  and 
to  the  three  rectilineal  figures,  viz.  the  squares  of  BD,  FH,  and 
the  square  which  is  equal  to  the  circle  ABCD,  there  is  a  fourth 
square  proportional;  because  to  the  three  straight  lines  which  are 
their  sides,  there  is  a  fourth  straight  line  proportional,  (12.  6.)  and 
this  fourth  square,  or  a  space  equal  to  it,  is  the  space  which  in 
this  proposition  is  denoted  by  the  letter  S:  and  the  like  is  to  be 
understood  in  the  other  places  above  cited;  and  it  is  probable  that 
this  has  been  shown  by  Euclid,  but  left  out  by  some  editor;  for 
the  lemma,  which  some  unskilful  hand  has  added  to  this  proposi- 
tion, explains  nothing  of  it. 

PROP.  III.     B.  XII. 

In  the  Greek  text  and  the  translations,  it  is  said,  "  and  because 
the  two  straight  lines  BA,  AC  which  meet  one  another,  "&c.;  here 
the  angles  BAC,  KHL  are  demonstrated  to  be  equal  to  one  ano- 
ther by  10th  prop.  B.  II.  which  had  been  done  before:  because  the 
triangle  EAG  was  proved  to  be  similar  to  the  triangle  KHL:  this 
repetition  is  left  out,  and  the  triaHgles  BAC,  KHL  are  proved  to 
be  similar  in  a  shorter  way  by  prop.  21,  B.  6. 

PROP.  IV.     B.  XII. 

A  few  things  in  this  are  more  fully  explained  than  in  the  Greek 
text. 

PROP.  V.     B.  XIL 

In  this,  near  to  the  end,  are  the  words  a^  e^Tr^os-^ev  ehtx^v)^  "as 
was  before  shown;"  and  the  same  are  found  again  in  the  end  of 
prop.  18,  of  this  book:  but  the  demonstration  referred  to,  except 
it  be  the  useless  lemma  annexed  to  the  2d  prop.,  is  no  where  in 
these  Elements,  and  has  been  perhaps  left  out  by  some  editor  who 
has  forgot  to  cancel  those  words  also. 


BOOK  XII. 


NOTES. 


287 


PROP.  VI.     B.  XII. 

A  shorter  demonstration  is  given  of  this;  and  that  which  is  ift 
the  Greek  text  may  be  made  shorter  by  a  step  than  it  is,  for  the 
authorof  it  makes  use  of  the  22d  prop,  of  B.  5,  twice;  whereas  once 
would  have  served  his  purpose:  because  that  proposition  extends 
to  any  number  of  magnitudes  which  are  proportionals  taken  two 
and  two,  as  well  as  to  three  which  are  proportional  to  other  three. 

COR.    PROP.  VIII.     B.  XII. 

The  demonstration  of  this  is  imperfect,  because  it  is  not  shown 
that  the  triangular  pyramids  into  which  those  upon  multangular 
bases  are  divided,  are  similar  to  one  another,  as  ought  necessarily 
to  have  been  done,  and  is  done  in  the  like  case  in  prop.  12th  of 
this  book.     The  full  demonstration  of  the  corollary  is  as  follows: 

Upon  the  polygonal  bases  ABCDE,  FGHKL,  let  there  be  simi- 
lar and  similai-ly  situated  pyramids  which  have  the  points  M,  N 
for  their  vertices:  the  pyramid  ABCDEM  has  to  the  pyramid 
FGHKLN  the  triplicate  ratio  of  that  which  the  side  AB  has  to 
the  homologous  side  FG. 

Let  the  polygons  be  divided  into  the  triangles  ABE,  EBC,  ECD; 
FGL,  LGH,  LHK,  which  are  similar  (20.  6.)  each  to  each,  and 
because  the  pyramids  are  similar,  therefore  (11.  def.  11.)  the  tri- 
angle EAM  is  similar  to  the  triangle  LFN,  and  the  triangle  ABM  to 
FGN:  wherefore  (4.  6.)  ME  is  to  EA,  as  NL  to  LF;  and  as  AE  to 
EBjSo  is  FL  to  LG,  because  the  triangles  EAB,LFG  are  similar; 
therefore,  ex  xquali^  as  ME  to  EB,  so  is  NL  to  LG:  in  like  man- 
ner it  may  be  shown  that  EB  is  to  BM,  as  LG  to  GN;  therefore 
again,  ex  aequali,  as  EM  to  MB,  so  is  LN  to  NG;  wherefore  the  tri- 

M 


N 


c/  v- 


angles  EMBjLNG  having  their  sides  proportionals,  are  (5.6.)  equi- 
angular and  similar  to  one  another:  therefore  the  pyramids  which 
have  the  triangles  EAB,  LEG  for  their  bases,  and  the  points  M, 
N  for  their  vertices,  are  similar  (l  I.  def.  11.)  to  one  another,  for 
their  solid  angles  are  (b.  11.)  equal,  and  the  solids  themselves  are 
contained  by  the  same  number  of  similar  planes:  in  the  same  man- 
ner, the  pyramid  EBCM  may  be  shown  to  be  similar  to  the  pyra- 
mid LGHN,  and  the  pyramid  ECDM  to  LHKN.  And  because 
the  pyramids  EABM,  LFGN  are  similar,  and   have  triangular 


288  NOTES.  BOOK  XII. 

bases,  the  pyramid  EABM  has  (8.  12.)  to  LFGN  the  triplicate 
ratio  of  that  which  EB  has  to  the  homologous  side  LG.     And,  in 
the  same  manner,  the  pyramid  EBCM  has  to  the  pyramid  LGHN 
the  triplicate  ratio  ot"  that  which  EB  has  to  LG.     Therefore  as 
the  pyramid  EABM  is  to  the  pyramid  LFGN,  so  is  the  pyramid 
EBCM  to  the  pyramid  LGHN.     In  like  manner,  as  the  pyramid 
EBCM  is  to  LGHN,  so  is  the  pyramid  ECDM  to  the  pyramid 
LHKM.     And  as  one  of  the  antecedents  is  to  one  of  the  conse- 
quents, so  are  all  the  antecedents  to  all  the  consequents:  therefore 
as  the  pyramid  EABM  to  the  pyramid  LFGN,  so  is  the  whole 
pyramid  ABCDEM  to  the  whole  pyramid  FGHKLN:   and  the 
pyramid  EABM  has  to  the  pyramid  LFGN  the  triplicate  ratio  of 
that  which  AB  has  to  YG;  therefore  the  whole  pyramid  has  to  the 
whole  pyramid  the  triplicate  ratio  of  that  which  AB  has  to  the 
homologous  side  FG.  Q.  E.  D.     , 

PROP.  XL  and  XH.     B.  XH. 

The  order  of  the  letters  of  the  alphabet  is  not  observed  in  these 
two  propositions  according  to  Euclid's  manner,  and  is  now  re- 
storedj  by  which  means,  the  first  part  of  prop.  12  may  be  demon- 
strated in  the  same  words  with  the  first  part  of  prop.  11:  on  this 
account  the  demonstration  of  that  first  part  is  left  out,  and  assumed 
from  prop.  11. 

PROP.  XH.   B.  xn. 

In  this  proposition,  the  common  section  of  a  plane  parallel  to 
the  bases  of  a  cylinder,  with  the  cylinder  itself,  is  supposed  to  be  a 
circle,  and  it  was  thought  proper  briefly  to  demonstrate  itj  from 
whence  it  is  sufficiently  manifest,  that  this  plane  divides  the  cy- 
linder into  two  others^  and  the  same  thing  is  understood  to  be 
supplied  in  prop.  14. 

PROP.  XV.     B.  XII. 

"And  complete  the  cylinders  AX,  EO,"  both  the  enunciation 
and  exposition  of  the  proposition  represent  the  cylinders  as  well 
as  the  cones,  as  already  described:  wherefore  the  reading  ought 
rather  to  be,  "  and  let  the  cones  be  ALC,  ENG^  and  the  cylinders 
AX,EO." 

The  first  case  in  the  second  part  of  the  demonstration  is  want- 
ing; and  something  also  in  the  second  case  of  that  part,  before  the 
repetition  of  the  construction  is  mentioned;  which  are  now  added. 

PROP.  XVII.     B.  XII. 

In  the  enunciation  of  this  proposition,  the  Greek  words  etg  niv 

f^tK^ovoc  a-Cpcii^uv  a-re^eov  "TroXveo^ov  eyf^ct'^xi  f/,^  i^ccvoy  rr^  eXxa-a-ovog  c(pecipxi 
Kuroc  7i}v  £7n(pu.veiccv  are  thus  translated  by  Commandine  and  others, 
"in  majori  solidum  polyhedrum  describere  quod  minoris  sphaerae 
superficiem  non  tangal;"  that  is, "  to  describe  in  the  greater  sphere 
a  solid  polyhedron  which  shall  not  meet  the  superficies  of  the 
lesser  sphere;"  whereby  they  refer  the  words  xcctcc  rt^v  s7ri<^xvetocv  to 
these  next  to  them  t>j§  eXcccra-ovo^  (r<poci^us.  But  they  ought  by  no 
means  to  be  thus  translated;  for  the  solid  polyhedron  doth  not 
only  meet  the  superficies  of  the  lesser  sphere,  but  pervades  the 


BOOK  XII.  NOTES.  289 

whole  of  that  spherei  therefore  the  aforesaid  words  are  to  be  re- 
ferred to  TO  a-Ts^eov  ^eAt/eJ'^ov,  and  ought  thus  to  be  translated,  viz. 
to  describe  in  the  greater  sphere  a  solid  polyhedron  whose  super- 
ficies shall  not  meet  the  lesser  sphere;  as  the  meaning  of  the  pro- 
position necessarily  requires. 

The  demonstration  of  the  proposition  is  spoiled  and  mutilated: 
for  some  easy  things  are  very  explicitly  demonstrated,  while  others 
not  so  obvious  are  not  sufficiently  explained:  for  example,  when 
it  is  affirmed,  that  the  square  of  KB  is  greater  than  the  double 
of  the  square  of  BZ,  in  the  first  demonstration,  and  that  the  angle 
BZK  is  obtuse,  in  the  second;  both  which  ought  to  have  been  de- 
monstrated. Besides,  in  the  first  demonstration  it  is  said,  "  draw 
Kq  from  the  point  K  perpendicular  to  BD;"  whereas  it  ought  to 
have  been  said  "join  KV,"  and  it  should  have  been  demonstrated 
that  KV  is  perpendicular  to  BD:  for  it  is  evident  from  the  figure 
in  Hervagius's  and  Gregory's  editions,  and  from  the  words  of  the 
demonstration,  that  the  Greek  editor  did  not  perceive  that  the 
perpendicular  drawn  from  the  point  K  to  the  straight  line  BD 
must  necessarily  fall  upon  the  point  V,  for  in  the  figure  it  is  made 
to  fall  upon  the  point  n,  a  different  point  from  V,  which  is  like- 
wise supposed  in  the  demonstration.  Commandine  seems  to  have 
been  aware  of  this:  for  in  this  figure  he  makes  one  and  the  same 
point  with  the  two  letters  V,  i2;  and  before  Commandine,  the 
learned  John  Dee,  in  the  commentary  he  annexes  to  this  proposi- 
tion in  Henry  Billinsley's  translation  of  the  Elements,  printed  at 
London,  ann.  1570,  expressly  takes  notice  of  this  error,  and  gives 
a  demonstration  suited  to  the  construction  in  the  Greek  text,  by 
which  he  shows  that  the  perpendicular  drawn  from  the  point  K 
to  BD,  must  necessarily  fall  upon  the  point  V. 

Likewise  it  is  not  demonstrated,  that  the  quadrilateral  figures 
SOFT,  TPRY,  and  the  triangle  YRX,  do  not  meet  the  lesser 
sphere,  as  was  necessary  to  have  been  done:  only  Clavius,  as  far 
as  I  know,  has  observed  this,  and  demonstrated  it  by  a  lemma, 
which  is  now  premised  to  this  proposition,  something  altered  and 
more  briefly  demonstrated. 

In  the  corollary  of  this  proposition,  it  is  supposed  that  a  solid 
polyhedron  is  described  in  the  other  sphere  similar  to  that  which 
is  described  in  the  sphere  BCDE;  but,  as  the  construction  by 
which  this  may  be  done  is  not  given,  it  was  thought  proper  to 
give  it,  and  to  demonstrate,  that  the  pyramids  in  it  are  similar  to 
those  of  the  same  order  in  the  solid  polyhedron  described  in  the 
sphere  BCDE. 

From  the  preceding  notes,  it  is  sufficiently  evident  how  much 
the  Elements  of  Euclid,  who  was  a  most  accurate  geometer,  have 
been  vitiated  and  mutilated  by  ignorant  editors.  The  opinion 
which  the  greatest  part  of  learned  men  have  entertained  concern- 
ing the  present  Greek  edition,  viz.  that  it  is  very  little  or  nothing 
different  from  the  genuine  Avork  of  Euclid,  has  without  doubt  de- 
ceived them,  and  made  them  less  attentive  and  accurate  in  exa- 
mining that  edition;  whereby  several  errors,  some  of  them  gross 
enough,  have  escaped  their  notice,  from  the  age  in  which  Theon 
2  O 


290  NOTES.  BOOK  XII. 

lived  to  this  time.  Upon  which  account  there  is  some  ground  to 
hope  that  the  pains  we  have  taken  in  correcting  those  errors,  and 
freeing  the  Elements  as  far  as  we  could  from  blemishes,  will  not 
be  unacceptable  to  good  judges,  who  can  discern  when  demon- 
strations are  legitimate,  and  when  they  are  not. 

The  objections  which,  since  the  first  edition,  have  been  made 
against  some  things  in  the  notes,  especially  against  the  doctrine 
of  proportionals,  have  either  been  fully  answered  in  Dr.  Barrow's 
Lect.  Mathemat.  and  in  these  notes;  or  are  such,  except  one 
which  has  been  taken  notice  of  in  the  note  on  prop.  1,  book  11,  as 
show  that  the  person  who  made  them  has  not  sufficiently  consi- 
dered the  things  against  which  they  are  brought;  so  that  it  is  not 
necessary  to  make  any  further  answer  to  these  objections  and 
others  like  them  against  Euclid's  definition  of  proportionals:  of 
which  definition  Dr.  Barrow  justly  says,  in  page  297  of  the  above 
named  book,  that  "Nisi  machinis  impulsa  validioribus  seternum 
persistet  inconcussa." 


FINIS. 


EUCLID'S  DATA. 


IN  THIS  EDITION, 


SEVERAL  ERRORS  ARE  CORRECTED, 


AND 


SOME  PROPOSITIONS  ADDED. 


BY  ROBERT  SIMSON,  M.  D. 

EMKRITtrS  PnOFESSOR  OF  MATHEMATICS  IX  THE  UNIVERSITY  OF  GLASGOW. 


PREFACE. 


■^■^  **^5  ^5  ©^♦^^■^ 


EUCLID'S  DATA  is  the  first  in  order  of  the  books 
written  by  the  ancient  geometers  to  facihtate  and  pro- 
mote tlie  method  of  resolution  or  analysis.  In  the  ge- 
neral, a  thing  is  said  to  be  given  which  is  either  ac- 
tually exhibited,  or  can  be  found  out,  that  is,  which  is 
cither  known  by  hypothesis,  or  that  can  be  demon- 
strated to  be  known;  and  the  propositions  in  the  book 
of  Euclid's  Data  show  what  things  can  be  found  out  or 
known  from  those  that  by  hypothesis  are  already 
known;  so  that  in  the  analysis  or  investigation  of  a 
problem,  from  the  things  that  are  laid  down  to  be 
known  or  given,  by  the  help  of  these  propositions  other 
things  are  demonstrated  to  be  given,  and  from  these, 
other  things  are  again  shown  to  be  given,  and  so  on, 
until  that  which  was  proposed  to  be  found  out  in  the 
problem  is  demonstrated  to  be  given,  and  when  this  is 
done,  the  problem  is  solved,  and  its  composition  is 
made  and  derived  from  the  compositions  of  the  Data 
which  were  made  use  of  in  the  analysis.  And  thus  the 
Data  of  Euclid  are  of  the  most  general  and  necessary 
use  in  the  solution  of  problems  of  every  kind. 

Euclid  is  reckoned  to  be  the  author  of  the  Book  of 
the  Data,  both  by  the  ancient  and  modern  geometers; 
and  there  seems  to  be  no  doubt  of  his  having  written  a 
book  on  this  subject,  but  which  in  the  course  of  so 
many  ages,  has  been  much  vitiated  by  unskilful  editors 
in  several  places,  both  in  the  order  of  the  propositions, 


294  PREFACE.  , 

and  in  the  definitions  and  demonstrations  themselves* 
To  correct  the  errors  which  are  now  found  in  it,  and 
bring  it  nearer  to  the  accuracy  with  which  it  was,  no 
doubt,  at  first  written  by  Euchd,  is  the  design  of  this 
edition,  that  so  it  may  be  rendered  more  useful  to  geo- 
meters, at  least  to  beginners  who  desire  to  learn  the 
investigatory  method  of  the  ancients.  And  for  their 
sakes,  the  compositions  of  most  of  the  Data  are  sub- 
joined to  their  demonstrations,  that  the  compositions 
of  problems  solved  by  help  of  the  Data  may  be  the 
more  easily  made. 

Marinus  the  philosopher's  preface,  which,  in  the 
Greek  editions,  is  prefixed  to  the  Data,  is  here  left  out, 
as  being  of  no  use  to  understand  them.  At  the  end  of 
it,  he  says,  that  Euclid  has  not  used  the  synthetical, 
but  the  analytical  method  in  delivering  them;  in  Avhich 
he  is  quite  mistaken;  for,  in  the  analysis  of  a  theorem, 
the  thing  to  be  demonstrated  is  assumed  in  the  analy- 
sis; but  in  the  demonstrations  of  the  Data,  the  thing 
to  be  demonstrated,  which  is,  that  something  or  other 
is  given,  is  never  once  assumed  in  the  demonstration, 
from  which  it  is  manifest,  that  every  one  of  them  is 
demonstrated  synthetically;  though,  indeed,  if  a  propo- 
sition of  the  Data  be  turned  into  a  problem,  for  exam- 
ple the  84th  or  85th  in  the  former  editions,  which  here 
are  the  85th  and  86th,  the  demonstration  of  the  propo- 
sition becomes  the  analysis  of  the  problem. 

Wherein  this  edition  diflfers  from  the  Greek,  and  the 
reasons  of  the  alterations  from  it,  will  be  shown  in  the 
notes  at  the  end  of  the  Data. 


EUCIilD'S  DATA. 


'  "^^Vf© ^5  *'^"'^" 


DUFINITIOirS. 


I. 

Spaces,  lines,  and  angles,  are  said  to  be  given  in  magnitude,  when 
equals  to  them  can  be  found. 

II. 

A  ratio  is  said  to  be  given,  vi^hen  a  ratio  of  a  given  magnitude  to 
a  given  magnitude  which  is  the  same  ratio  with  it  can  be  found. 

III. 

Rectilineal  figures  are  said  to  be  given  in  species,  which  have 

each  of  their  angles  given,  and  the  ratios  of  their  sides  given. 

IV. 

Points,  lines,  and  spaces,  are  said  to  be  given  in  position,  which 
have  always  the  same  situation,  and  which  are  either  actually 
exhibited,  or  can  be  found. 

A. 

An  angle  is  said  to  be  given  in  position,  which  is  contained  by 
straight  lines  given  in  position. 

V. 

A  circle  is  said  to  be  given  in  magnitude,  when  a  straight  line 
from  its  centre  to  the  circumference  is  given  in  magnitude. 

VI. 

A  circle  is  said  to  be  given  in  position  and  magnitude,  the  centre 
of  which  is  given  in  position,  and  a  straight  line  from  it  to  the 
circumference  is  given  in  magnitude. 

VII. 
Segments  of  circles  are  said  to  be  given  in  magnitude,  when  the 
angles  in  them,  and  their  bases,  are  given  in  magnitude. 

VIII. 
Segments  of  circles  are  said  to  be  given  in  position  and  magni- 
tude, when  the  angles  in  them  are  given  in  magnitude,  and  their 
bases  are  given  both  in  position  and  magnitude. 

IX. 

A  magnitude  is  said  to  be  greater  than  another  by  a  given  mag- 
nitude, when  this  given  magnitude  being  taken  from  it,  the  re- 
mainder is  equal  to  the  other  magnitude. 

X. 

A  magnitude  is  said  to  be  less  than  another  by  a  given  magni- 
tude, when  this  given  magnitude  being  added  to  it,  the  whole 
is  equal  to  the  other  magnitude. 


296 


EUCLID  S  DATA. 


PROPOSITION  I.  *1. 

The  ratios  of  a^iven  magnitudes  to  one  another  are 
given.t 

Let  Aj  B  be  two  given  magnitudes,  the  ratio  of  A  to  B  is  given. 

Because  A  is  a  given  magnitude,  there  may 
(1.  def.  Dat.)  be  found  one  equal  to  itj  let  this 
be  C,  and  because  B  is  given,  one  equal  to  it 
maybe  foundj  let  it  be  D;  and  since  A  is  equal 
to  C,  and  B  to  B;  therefore  (7.  5.)  A  is  to  B, 
as  C  to  D5  and  consequently  the  ratio  of  A  to 
B  is  given,  because  the  ratio  of  the  given  mag- 
nitudes C,  D,  which  is  the  same  with  it,  has  A  B  C  D 
been  found. 


PROP.  II. 


2. 


ABC 


E 


If  a  given  magnitude  has  a  given  ratio  to  another 
magnitude,  "  and  if  unto  the  two  magnitudes,  by  which 
the  given  ratio  is  exhibited,  and  the  given  magnitude,  a 
fourth  proportional' can  be  found;"  the  other  magni- 
tude is  given.t 

Let  the  given  magnitude  A  have  a  given  ratio  to  the  m-agni- 
tude  B;  if  a  fourth  proportional  can  be  found  to  the  three  magni- 
tudes above  named,  B  is  given  in  magnitude. 

Because  A  is  given,  a  magnitude  may  be 
found  equal  to  it  (l.  def.);  let  this  be  C:  and 
because  the  ratio  of  A  to  B  is  given,  a  ratio 
which  is  the  same  with  it  may  be  found;  let 
this  be  the  ratio  of  the  given  magnitude  E  to 
the  given  magnitude  F:  unto  the  magnitudes 
E,  F,  C  find  a  fourth  proportional  D,  which,  by 
the  hypothesis,  can  be  done.  Wherefore  be- 
cause A  is  to  B,  as  E  to  F;  and  as  E  to  F,  so 
is  C  to  D;  A  is  (11.  5.)  to  B,  as  C  to  D.  But 
A  is  equal  to  C;  therefore  (14.  5.)  B  is  equal  to  D.  The  magni- 
tude B  is  therefore  given  (1.  def.)  because  a  magnilude  D  ecjual 
to  it  has  been  found. 

The  limitation  within  the  inverted  commas  is  not  in  the  Greek 
text,  but  is  now  necessarily  added;  and  the  same  must  be  under- 
stood in  all  the  propositions  of  the  book  which  depend  upon  this 
second  proposition,  where  it  is  not  expressly  mentioned.  See  the 
note  upon  it. 


*  The  figures  in  the  margin  show  the  number  of  the  propositions  in  the  other 
editions. 
f  See  Note. 


Euclid's  data.  297 

PROP.  III.  3. 

If  any  given  magnitudes  be  added  together,  their  sum 
shall  be  given. 

Let  any  given  magnitudes  AB,  BC  be  added  together,  their 
sum  AC  is  given. 

Because  AB  is  given,  a  magnitude  equal  to  it  may  be  found 
(l.  def.);  let  this  be  DE:  and  because     ^  B  C 

I3C  is  given,  one  equal  to  it  may  be i 

found;   let  this  be  EF:    wherefore,  be- 
cause AB  is  equal  to  DE,  and  BC  equal     jy  E  F 

to  EF|   the  whole  AC  is  equal  to  the i 

whole  DF:  AC  is  therefore  given,  be-  ' 

cause  DF  has  been  found,  which  is  equal  to  it. 

PROP.  IV.  4. 

If  a  given  magnitude  be  taken  from  a  given  magni- 
tude, the  remaining  magnitude  shall  be  given. 

From  the  given  magnitude  AB,  let  the  given  magnitude  AC 
be  taken;  the  remaining  magnitude  CB  is  given. 

Because  AB  is  given,  a  magnitude  equal  to  it  may  (l.  def.)  be 
found;  let  this  be  DE:   and  because        a  C  B 

AC  is  given,  one  equal  to  it  may  be       J I ^ 

found;    let   this    be    DF:    wherefore  ' 

because  AB  is  equal  to  DEj  and  AC       j-j  F  E 

to  DF;    the  remainder  CB    is  equal       ^^ i 

to  the  remainder  FE.    CB  is  therefore 

given  (1.  def.),  because  FE,  which  is  equal  to  it,  has  been  found. 

PROP.  V.  12. 

If  of  three  magnitudes,  the  first  together  with  the  se- 
cond be  given,  and  also  the  second  together  with  the 
third;  either  the  first  is  equal  to  the  third,  or  one  of  them 
is  greater  than  the  other  by  a  given  magnitude.* 

Let  AB,  BC,  CD  be  three  magnitudes,  of  which  AB  together 
with  BC,  that  is  AC,  is  given;  and  also  BC  together  with  CD, 
that  is,  BD,  is  given.  Either  AB  is  equal  to  CD,  or  one  of  them 
is  greater  than  the  other  by  a  given  magnitude. 

Because  AC,   BD   are  each   of  them   given,  they   are  either 
equal  to  one  another,  or  not  equal.   First, 
let  them  be  equal,   and   because  AC  is       ^     g  CD 

equal  to  BD,  take  away  the  common  part       i \ 

BC;  therefore  the  remainder  AB  is  equal 
to  the  remainder  CD. 

But  if  they  be  unequal,  let  AC  be  greater  than  BD,  and  make 
CE  equal  to  BD.     Therefore  CE  is  given,  because  BD  is  given. 
And  the  whole  AC  is  given;  there- 
fore (4,  dat.)   AE    the  remainder  is       A     E  B  CD 
given.     And  because  EC  is  equal  to       - — 

•*  See  Note. 
2P 


298  Euclid's  data. 

BD,  by  taking  BC  from  both,  the  remainder  EB  is  equal  to  the 
remainder  CD.  And  AE  is  givenj  wherefore  AB  exceeds  EB, 
that  is  CD,  by  the  given  magnitude  AE. 

PROP.  VI.  5. 

If  a  magnitude  has  a  given  ratio  to  a  part  of  it,  it 
shall  also  have  a  given  ratio  to  the  remaining  part  of  it.* 

Let  the  magnitude  AB  have  a  given  ratio  to  AC  a  part  of  it; 
it  has  also  a  given  ratio  to  the  remainder  BC. 

Because  the  ratio  of  AB  to  AC  is  given,  a  ratio  may  be  found 
(2.  def.)  which  is  the  same  to  it:  let  this  be  the  ratio  of  DE  a 
given  magnitude  to  the  given  magnitude      »  C  B 

DF.     And  because  DE,  DF  are  given. 


the  remainder  FE  is  (4.  dat.)  given:  and     y.  F  E 

because  AB  is  to  AC,  as  DE  to  DF,  by , 

conversion  (E.  5.)  AB  is  to  BC,  as  DE  ' 

to  EF.  Therefore  the  ratio  of  AB  to  BC  is  given,  because  the 
ratio  of  the  given  magnitudes  DE,  EF,  which  is  the  same  with  it, 
has  been  found. 

CoR.  From  this  it  follows,  that  the  parts  AC,  CB  have  a  given 
ratio  to  one  another:  because  as  AB  to  BC,  so  is  DE  to  EF; 
by  division  (17.  5.),  AC  is  to  CB,  as  DF  to  FE:  and  DF,  FE  are 
given;  therefore  (2.  def.)  the  ratio  of  AC  to  CB  is  given. 

PROP.  VII.  6. 

If  two  magnitudes  which  have  a  given  ratio  to  one 
another,  be  added  together;  the  whole  magnitude  shall 
have  to  each  of  them  a  given  ratio.* 

Let  the  magnitudes  AB,  BC  which  have  a  given  ratio  to  one 
another,  be  added  together;  the  whole  AC  has  to  each  of  the 
magnitudes  AB,  BC  a  given  ratio. 

Because  the  ratio  of  AB  to  BC  is  given,  a  ratio  may  be  found 
(2.  def.)  which  is  the  same  with  it;  let  this  be  the  ratio  of  the 
given  magnitudes  DE,  EF:   and  because      .  B  C 

DE,  EF  are  given,  the  whole  DF  is  given . 

(3.  dat.):  and  because  as  AB  to  BC,  so  is  ' 

DE  to  EF;  by  composition  (18.  5.),  AC  is     j^  F         F 

to  CB,  as  DF  to  FE;  and  by  conversion  (E.     i 

5.),  AC  is  to  AB,  as  DF  to  DE;  wherefore  ' 

because  AC  is  to  each  of  the  magnitudes  AB,  BC,  as  DF  to  each 
of  the  others  DE,  EF;  the  ratio  of  AC  to  each  of  the  magnitudes 
AB,  BC  is  given  (2.  def.). 

PROP.  VIII.  7. 

If  the  given  magnitude  be  divided  into  two  parts 
which  have  a  given  ratio  to  one  another,  and  if  a  fourth 

"  See  Note. 


EUCLID  S  DATA. 


299 


proportional  can  be  found  to  the  sum  of  the  two  mag- 
nitudes by  which  the  given  ratio  is  exhibited,  one  of 
them,  and  the  given  magnitude;  each  of  the  parts  is 
given.* 

Let  the  given  magnitude  AB  be  divided  into  the  parts  AC,  CB 
which  have  a  given  ratio  to  one  another;  if  a  fourth  proportional 
can  be  found  to  the  above  named  magni-      .  r  Ti 

tudes;  AC  and  CB  are  each  of  them  given. 

Because  the  ratio  of  AC  to  CB  is  given, 
the  ratio  of  AB  to  BC  is  given  (7.  dat.); 
therefore  a  ratio  which  is  the  same  with  it 
can  be  found  (2.  def.);  let  this  be  the  ratio 
of  the  given  magnitudes,  DE,  EF:  and  be- 
cause the  given  magnitude  AB  has  to  BC 
the  given  ratio  of  DE  to  EF,  if  unto  DE, 
EF,  AB  a  fourth  proportional  can  be 
found,  this  which  is  BC  is  given  (2.  dat.); 
and  because  AB  is  given,  the  other  part  ' 

AC  is  given  (4.  dat.) 

In  the  same  manner,  and  with  the  like  limitation,  if  the  differ- 
ence AC  of  two  magnitudes  AB,  BC  which  have  a  given  ratio  be 
given;  each  of  the  magnitudes  AB,  BC  is  given. 


D 


A 


D 


C 


B 


F 


E 


PROP.  IX. 


8. 


Magnitudes  which  have  given  ratios  to  the  same 
magnitude,  have  also  a  given  ratio  to  one  another. 

Let  A,  C  have  each  of  them  a  given  ratio  to  B;  A  has  a  given 
ratio  to  C. 

Because  the  ratio  of  A  to  B  is  given,  a  ratio  which  is  the  same 
to  it  may  be  found  (2.  def.);  let  this  be  the  ratio  of  the  given  mag- 
nitudes D,  E:  and  because  the  ratio  of  B  to  C  is  given,  a  ratio 
which  is  the  same  with  it  may  be  found  (2.  def.);  let  this  be  the 
ratio  of  the  given  magnitudes  F,  G: 
to  F,  G,  E  find  a  fourth  proportional 
H,  if  it  can  be  done;  and  because 
as  A  is  to  B,  so  is  D  to  E;  and  as  B 
to  C,  so  is  (F  to  G,  and  so  is)  E  to 
H;  ex  eequalij  as  A  to  C,  so  is  D  to 
H:  therefore  the  ratio  of  A  to  C  is  A  B  C 
given  (2.  def.)  because  the  ratio  of 
the  given  magnitudes  D  and  H, 
which  is  the  same  with  it,  has  been 
found:  but  if  a  fourth  proportional 
to  F,  G,  E  cannot  be  found,  then  it 
can  only  be  said  that  the  ratio  of  A  to  C  is  compounded  of  the 
ratios  of  A  to  B,  and  B  to  C,  that  is,  of  the  sriven  ratios  of  D  to  E, 
and  F  to  G. 


D      E 
F 


H 
G 


*  See  Note. 


300  Euclid's  data. 

PROP.  X.  9. 

If  two  or  more  magnitudes  have  given  ratios  to  one 
another,  and  if  they  have  given  ratios,  though  they  be 
not  the  same,  to  some  other  magnitudes;  these  other 
magnitudes  shall  also  have  given  ratios  to  one  another. 

Let  two  or  more  magnitudes  A,  B,  C  have  given  ratios  to  one 
another;  and  let  them  have  given  ratios,  though  they  be  not  the 
same,  to  some  other  magnitudes  D,  E,  F;  the  magnitudes  D,  E, 
F  have  given  ratios  to  one  another. 

Because  the  ratio  of  A  to  B  is  given,  and  likewise  the  ratio  of 
A  to  D;  therefore  the  ratio  of  D  to 

B  is  given  (9.  dat.):  but  the  rati/)  of    A D 

B  to  E  is  given,  therefore  (9.  dat.)     B E — — • 

the  ratio  of  D  to  E  is  given:  and  be-     C F 

cause  the  ratio  of  B  to  C  is  given, 

and  also  the  ratio  of  B  to  E;  the  ratio  of  E  to  C  is  given  (9.  dat.): 
and  the  ratio  of  C  to  F  is  given;  wherefore  the  ratio  of  E  to  F  is 
given;  D,  E,  F  have  therefore  given  ratios  to  one  another. 

PROP.  XI.  22. 

If  two  magnitudes  have  each  of  them  a  given  ratio 
to  another  magnitude,  both  of  them  together  shall  have 
a  given  ratio  to  that  other. 

Let  the  magnitudes  AB,  BC  have  a  given  ratio  to  the  magni- 
tude D;  AC  has  a  given  ratio  to  the  same  D. 

Because  AB,  BC,  have  each  of  them  -d 

a  given  ratio  to  D,  the  ratio  of  AB  to      » ■ q 

BC  is  given  (9.  dat.):  and  by  composi-  ' 

tion,  the  ratio  of  AC  to  CB  is  given;     j^         

(7.  dat.):  but  the  ratio  of  BC  to  D  is 

given;  therefore  (9.  dat.)  the  ratio  of  AC  to  D  is  given. 

PROP.  Xn.  23. 

If  the  whole  have  to  the  whole  a  given  ratio,  and  the 
parts  have  to  the  parts  given,  but  not  the  same,  ratios, 
every  one  of  them,  whole  or  part,  shall  have  to  every 
one  a  given  ratio.* 

Let  the  whole  AB  have  a  given  ratio  to  the  whole  CD,  and  the 
parts  AE,  EB  have  given,  but  not  the  same,  ratios  to  the  parts 
CF,  FD,  every  one  shall  have  to  every  one,  whole  or  part,  a  given 
ratio. 

Because  the  ratio  of  AE  to  CF  is  given,  as  AE  to  CF,  so 
make  AB  to  CG;  the  ratio  therefore  of  AB  to  CG  is  given; 
wherefore  the  ratio  of  the  remainder  EB  to  the  remainder  FG  is 

*  See  Note. 


EUCLID  S  DATA. 


301 


given,  because  it  is  the  same  (l9.  5.)  with  the  ratio  of  AB  to  CG: 
and  the  ratio  of  EB  to  FD  is  given,  where-     a  E  B 

fore  the  ratio  of  FD  to  FG  is  given  (9. , 

dat.)^  and  by  conversion,  the  ratio  of  FD  ' 

to  DG  is  given  (6.  dat.):  and  because  AB     ^  p  CD 

has  to  each  of  the  magnitudes  CD,  CG  a     , , 

given  ratio,  the   ratio  of  GD   to  CG  is  '  ' 

given  (9.  dat.)^  and  therefore  (6.  dat.)  the  ratio  of  CD  to  DG  is 
given:  but  the  ratio  of  GD  to  DF  is  given,  wherefore  (9.  dat.)  the 
ratio  of  CD  to  DF  is  given,  and  consequently  (cor.  6.  dat.)  the 
ratio  of  CF  to  FD  is  given;  but  the  ratio  of  CF  to  AE  is  given, 
as  also  the  ratio  of  FD  to  EB,  wherefore  (10.  dat.)  the  ratio  of  AE 
to  EB  is  given;  as  also  the  ratio  of  AB  to  each  of  them  (7.  dat.): 
the  ratio  therefore  of  one  to  every  one  is  given. 


PROP.  XIII, 


24. 


If  the  first  of  three  proportional  straight  hnes  has  a 
given  ratio  to  the  third,  the  first  shall  also  have  a  given 
ratio  to  the  second.* 

Let  A,  B,  C  be  three  proportional  straight  lines,  that  is,  as  A 
to  B,  so  is  B  to  C;  if  A  has  to  C  a  given  ratio,  A  shall  also  have 
to  B  a  given  ratio. 

Because  the  ratio  of  A  to  C  is  given,  a  ratio  which  is  the  same 
with  it  may  be  found  (2.  def.);  let  this  be  the  ratio  of  the  given 
straight  lines  D,  E;  and  between  D  and  E  find  a  (13.  6.)  mean 
proportional  F;  therefore  the  rectangle  contained  by  D  and  E  is 
equal  to  the  square  of  F,  and  the  rectangle  D,  E 
is  given,  because  its  sides  D,  E  are  given;  where- 
fore the  square  of  F,  and  the  straight  line  F  is 
given:  and  because  as  A  is  to  C,  so  is  D  to  E; 
but  as  A  to  C,  so  is  (2.  cor.  20.  6.)  the  square  of 
A  to  the  square  of  B;  and  as  D  to  E,  so  is  (2.  cor. 
20.  6.)  the  square  of  D  to  the  square  of  F:  there- 
fore the  sqiiare  (11.  5.)  of  A  is  to  the  square  of  B, 
as  the  square  of  D  to  the  square  of  F:  as  there- 
fore (22.  6.)  the  straight  line  A  to  the  straight 
line  B,  so  is  the  straight  line  D  to  the  straight 
line  F:  therefore  the  ratio  of  A  to  B  is  given  (2. 
def.),  because  the  ratio  of  the  given  straight  lines 
D,  F,  which  is  the  same  with  it,  has  been  found. 

PROP.  XIV.  A. 

If  a  magnitude  together  with  a  given  magnitude  has 
a  given  ratio  to  another  magnitude;  the  excess  of  this 
other  magnitude  above  a  given  magnitude  has  a  given 
ratio  to  the  first  magnitude:  and  if  the  excess  of  a 
magnitude  above  a  given  magnitude  has  a  given  ratio 

*  See  Note. 


D 


E 
I 


302  Euclid's  data. 

to  another  magnitude;  this  other  magnitude  together 
with  a  given  magnitude  has  a  given  ratio  to  the  first 
magnitude.* 

Let  the  magnitude  AB  together  with  the  given  magnitude  BE, 
that  is,  AE,  have  a  given  ratio  to  the  magnitude  CD;  the  excess 
of  CD  above  a  given  magnitude  has  a  given  ratio  to  AB. 

Because  the  ratio  of  AE  to  CD  is  given,  as  AE  to  CD,  so 
make  BE  to  FD;  therefore  the  ratio  of  BE  to  FD  is  given,  and 
BE  is  given;  wherefore  FD  is   given      .  R  F 

(2.  dat.):  and  because  as  AE  to  CD,     _^___ 

so  is  BE  to  FD,  the  remainder  AB  is  » 

(19.  5.)  to  the  remainder  CF,  as  AE     C  F       D 

to  CD:  but  the  ratio  of  AE  to  CD  is     1 

given,  therefore  the  ratio  of  AB  to  CF  is  given;  that  is,  CF,  the 
excess  of  CD  above  the  given  magnitude  FD,  has  a  given  ratio 
to  AB. 

Next,  Let  the  excess  of  the  magnitude  AB  above  the  given 
magnitude  BE,  that  is,  let  AE  have  a  given  ratio  to  the  magni- 
tude CD:  CD  together  with  a  given  magnitude  has  a  given  ratio 
to  AB. 

Because  the  ratio  of  AE  to  CD  is  given,  as  AE  to  CD,  so 
make  BE  to  FD;  therefore  tlie  ratio  of     .  F     B 

BE  to  FD  is  given,    and  BE  is  given, 

wherefore  FD  is  given   (2.  dat.).     And         ~         "  ' 

because  as  AE  to  CD,  so  is  BE  to  FD,      C  D     F 

AB  is  to  CF,  as  (13.  5.)  AE  to  CD:  but 1 

the  ratio  of  AE  to  CD  is  given,  therefore  the  ratio  of  AB  to  CF 
is  given:  that  is,  CF,  which  is  equal  to  CD  together  with  the 
given  magnitude  DF,  has  a  given  ratio  to  AB. 

PROP.  XV.  B. 

If  a  magnitude,  together  with  that  to  which  another 
magnitude  has  a  given  ratio,  be  given;  the  sum  of  this 
other,  and  that  to  which  the  first  magnitude  has  a  given 
ratio,  is  given.* 

Let  AB,  CD  be  two  magnitudes,  of  which  AB  together  with 
BE,  to  which  CD  has  a  given  ratio,  is  given;  CD  is  given,  to- 
gether with  that  magnitude  to  which  AB  has  a  given  ratio. 

Because  the  ratio  of  CD  to  BE  is  given,  as  BE  to  CD,  so  make 
AE  to  FD;  therefore  the  ratio  of  AE  to  FD  is  given,  and  AE  is 
given,  wherefore  (2.  dat.)  FD  is  given:  ^  "R  F 
and  because  as  BE  to  CD,  so  is  AE  to  ■  


FD:  AB  is  (cor.  19.  5.)  to  EC,  as  BE  to 

CD:  and  the  ratio  of  BE  to  CD  is  given,     F  CD 

wherefore   the    ratio   of  AB    to   FC   is 1 

given:  and  FD  is  given,  that  is  CD  together  with  FC,  to  which 
AB  has  a  given  ratio,  is  given. 


*  See  Note. 


Euclid's  data,  303 

PROP.  XVI.  10. 

If  the  excess  of  a  magnitude,  above  a  given  magni- 
tude, has  a  given  ratio  to  another  magnitude;  the  ex- 
cess of  both  together  above  a  given  magnitude  shall 
have  to  that  other  a  given  ratio:  and  if  the  excess  of 
two  magnitudes  together  above  a  given  magnitude,  has 
to  one  of  them  a  given  ratio;  either  the  excess  of  the 
other  above  a  given  magnitude  has  to  that  one  a  given 
ratio,  or  the  other  is  given  together  with  the  magnitude 
to  which  that  one  has  a  given  ratio.* 

Let  the  excess  of  the  magnitude  AB  above  a  given  magnitude, 
have  a  given  ratio  to  the  magnitude  BC;  the  excess  of  AC,  both 
of  them  together,  above  the  given  magnitude,  has  a  given  ratio 
to  BC. 

Let  AD  be  the  given  magnitude,  the  excess  of  AB  above 
which,  viz.  DB,  has  a  given  ratio  to      .  n       R  C 

BC:  and  because  DB  has  a  given  ra- . 

tio  to  BC,  the  ratio  of  DC  to   CB  is  '         ' 

given  (7.  dat.),  and  AD  is  given;  therefore  DC,  the  excess  of 

AC  above  the  given  magnitude  AD,  has  a  given  ratio  to  BC. 

Next,  Let  the  excess  of  two  magnitudes  AB,  BC  together, 
above  a  given  magnitude,  have  to  one      .  D       "R       F     r 

of  them  BC  a  given  ratio;  either  the . 

excess  of  the  other  of  them  AB  above  '         '         ' 

the  given  magnitude  shall  have  to  BC  a  given  ratio;  or  AB  is 
given,  together  with  the  magnitude  to  which  BC  has  a  given 
ratio. 

Let  AD  be  the  given  magnitude,  and  first  let  it  be  less  than 
AB;  and  because  DC,  the  excess  of  AC  above  AD  has  a  given 
ratio  to  BC,  DB  has  (cor.  6.  dat.)  a  given  ratio  to  BC;  that  is, 
DB,  the  excess  of  AB  above  the  given  magnitude  AD,  has  a 
given  ratio  to  BC. 

But  let  the  given  magnitude  be  greater  than  AB,  and  make  AE 
equal  to  it;  and  because  EC,  the  excess  of  AC  above  AE,  has  to 
BC  a  given  ratio,  BC  has  (6.  dat.)  a  given  ratio  to  BE;  and  be- 
cause AE  is  given,  AB  together  with  BE,  to  which  BC  has  a 
given  ratio,  is  given. 

PROP.  XVn.  11. 

If  the  excess  of  a  magnitude  above  a  given  magni- 
tude have  a  given  ratio  to  another  magnitude;  the  ex- 
cess of  the  same  first  magnitude  above  a  given  magni- 
tude, shall  have  a  given  ratio  to  both  the  magnitudes 
together.  And  if  the  excess  of  either  of  two  magni- 
tudes above  a  given  magnitude  have  a  given  ratio  to 
both  magnitudes    together;    the   excess  of  the  same 

^  Sec  Note. 


304  Euclid's  data. 

above  a  given  magnitude  shall  have  a  given  ratio  to 
the  other. ^ 

Let  the  excess  of  the  magnitude  AB  above  a  given  magnitude 
have  a  given  ratio  to  the  magnitude  BC:  the  excess  of  AB  above 
a  given  magnitude  has  a  given  ratio  to  AC. 

Let  AD  be  the  given  magnitude^  and  because  DB,  the  excess 
of  AB  above  AD,  has  a  given  ratio  to  BC;  the  ratio  of  DC  to 
DB  is  given  (7.  dat.):  make  the  ratio  of  AD  to  DE  the  same 
with  this  ratio;  therefore  the  ratio 
of  AD  to  DE  is  given:  and    AD    is     A  E       D       B     C 

given,    wherefore   (2.   dat.)   DE,   and     1 1 1 

the  remainder  AE  are  given:  and  because  as  DC  to  DB,  so  is 
AD  to  DE,  AC  is  (12.  5.)  to  EB,  as  DC  to  DB;  and  the  ratio  of 
DC  to  DB  is  given;  wherefore  the  ratio  of  AC  to  EB  is  given: 
and  because  the  ratio  of  EB  to  AC  is  given,  and  that  AE  is 
given,  therefore  EB,  the  excess  of  AB  above  the  given  magni- 
tude AE,  has  a  given  ratio  to  AC. 

Next,  Let  the  excess  of  AB  above  a  given  magnitude  have  a 
given  ratio  to  AB  and  BC  together,  that  is,  to  AC;  the  excess  of 
AB  above  a  given  magnitude  has  a  given  ratio  to  BC. 

Let  AE  be  the  given  magnitude;  and  because  EB,  the  excess 
of  AB  above  AE  has  to  AC  a  given  ratio,  as  AC  to  EB,  so  make 
AD  to  DE;  therefore  the  ratio  of  AD  to  DE  is  given,  as  also 
(6.  dat.)  the  ratio  of  AD  to  AE:  and  AE  is  given,  wherefore 
(2.  dat.)  AD  is  given:  and  because,  as  the  whole  AC,  to  the 
whole  EB,  so  is  AD  to  DE,  the  remainder  DC  is  (19.  5.)  to  the 
remainder  DB,  as  AC  to  EB;  and  the  ratio  of  AC  to  EB  is 
given;  wherefore  the  ratio  of  DC  to  DB  is  given,  as  also  (cor.  6. 
dat.)  the  ratio  of  DB  to  BC:  and  AD  is  given;  therefore  DB,  the 
excess  of  AB  above  a  given  magnitude  AD,  has  a  given  ratio  to 
BC. 

PROP.  XVIIL  14. 

If  to  each  of  two  magnitudes,  which  have  a  given 
ratio  to  one  another,  a  given  magnitude  be  added;  the 
wholes  shall  either  have  a  given  ratio  to  one  another,  or 
the  excess  of  one  of  them  above  a  given  magnitude  shall 
have  a  given  ratio  to  the  other. 

Let  the  two  magnitudes  AB,  CD  have  a  given  ratio  to  one  an- 
other, and  to  AB  let  the  given  magnitude  BE  be  added,  and  the 
given  magnitude  DF  to  CD:  the  wholes  AE,  CF  either  have  a 
given  ratio  to  one  another,  or  the  excess  of  one  of  them  above  a 
given  magnitude  has  a  given  ratio  to  the  other  (1.  dat.). 

Because  BE,  DF  are  each  of  them  given,  their  ratio  is  given, 
and  if  this  ratio  be  the  same  with     A         B  E 

the  ratio  of  AB  to  CD,  the  ratio  of    1 

AE  to  CF,  which  is  the  same  (12.  5.) 

with  the  given  ratio  of  AB  to  CD,     CD  F 

shall  be  given.  1 -— 

^  See  Note, 


Euclid's  data.  305 


But  if  the  ratio  of  BE  to  DF  be  not  the  same  with  the  ratio  of 
AB  to  CD,  either  it  is  ,^reater  than  the  ratio  of  AB  to  CD,  or,  by- 
inversion,  the  ratio  of  DF  to  BE  is  greater  than  the  ratio  of  CD 
to  AB:  first,  let  the  ratio  of  BE  to  DF     A     B  G         E 

be  greater  than  the  ratio  of  AB  to  CD;     1 1 

and  as  AB  to  CD,  so  make  BG  to  DF; 

therefore  the  ratio  of  BG  to  DF  is  given;     CD  F 

and  DF  is  given,  therefore  (2.  dat.)  BG  is      1 

given:  and  because  BE  has  a  greater  ratio  to  DF  than  (AB  to 
CD,  that  is,  than)  BG  to  DF,  BE  is  greater  (10.  5.)  than  BG;  and 
because  as  AB  to  CD,  so  is  BG  to  DF;  therefore  AG  is  (12.  5.) 
to  CF,  as  AB  to  CD:  but  the  ratio  of  AB  to  CD  is  given,  where- 
fore the  ratio  of  AG  to  CF  is  given;  and  because  BE,  BG  are  each 
of  them  given,  GE  is  given:  therefore  AG,  the  excess  of  AE 
above  a  given  magnitude  GE,  has  a  given  ratio  to  CF.  The  other 
case  is  demonstrated  in  the  same  manner. 

PROP.  XIX.  15. 

If  from  each  of  two  magnitudes,  which  have  a  given 
ratio  to  one  another,  a  given  magnitude  be  taken,  the 
remainders  shall  either  have  a  given  ratio  to  one  ano- 
ther, or  the  excess  of  one  of  them  above  a  given  mag- 
nitude, shall  have  a  given  ratio  to  the  other. 

Let  the  magnitudes  AB,  CD  have  a  given  ratio  to  one  another, 
and  from  AB  let  the  given  magnitude  AE  be  taken,  and  from  CD, 
the  given  magnitude  CF:  the  remainders  EB,  FD  shall  either 
have  a  given  ratio  to  one  another,  or  the  excess  of  one  of  them 
above  a  given  magnitude  shall  have  a 
given  ratio  to  the  other.  A         E  B 

Because  AE,  CF  are  each  of  them  given,    1 

their  ratio  is  given  (l.  dat.):  and  if  this 

ratio  be  the  same  with  the  ratio  of  AB  to     C      F  D 

CD,  the  ratio  of  the  remainder  EB  to  the     


remainder  FD,  which  is  the  same  (19.  5.)  with  the  given  ratio  of 
AB  to  CD,  shall  be  given. 

But  if  the  ratio  of  AB  to  CD  be  not  the  same  with  the  ratio  of 
AE  to  CF,  either  it  is  greater  than  the  ratio  of  AE  to  CF,  or,  by 
inversion,  the  ratio  of  CD  to  AB  is  greater  than  the  ratio  of  CF 
to  AE.  First,  let  the  ratio  of  AB  to  CD  be  greater  than  the  ratio 
of  AE  to  CF,  and  as  AB  to  CD,  so  make  AG  to  CF;  therefore 
the  ratio  of  AG  to  CF  is  given,  and 
CF  is  given,  wherefore  (2.  dat.)  AG  is     A  EG  B 

given:  and  because  the  ratio  of  AB  to 1 — | 

CD,  that  is,  the  ratio  of  AG  to  CF,  is 

greater  than   the  ratio  of  AE  to  CF;     C  F     D 

AG  is  greater  (10.  5.)  than  AE:  and 1 

AG,  AE  are  given,  therefore  the  remainder  EG  is  given;  and  as 
AB  to  CD,  so  is  AG  to  CF,  and  so  is  (19.  5.)  the  remainder  GB 
to  the  remainder  FD;  and  the  ratio  of  AB  to  CD  is  given:  where- 
fore the  ratio  of  GB  to  FD  is  given;  therefore  GB,  the  excess  of 
2  Q 


306  Euclid's  data,  •' 

EB  above  a  given  magnitude  EG,  has  a  given  ratio  to  FD.     Its 
the  same  manner  the  other  case  is  demonstrated. 

PROP.  XX.  16. 

If  to  one  of  two  magnitudes  which  have  a  given 
ratio  to  one  another,  a  given  magnitude  be  added,  and 
from  the  other  a  given  magnitude  be  taken;  the  excess 
of  the  sum  above  a  given  magnitude  shall  have  a  given 
ratio  to  the  remainder. 

Let  the  two  magnitudes  AB,  CD  have  a  given  ratio  to  one  ano- 
ther, and  to  AB  let  the  given  magnitude  EA  be  added,  and  from 
CD  let  tlie  given  magnitude  CF  be  takenj  the  excess  of  the  sum. 
EB  above  a  given  magnitude,  has  a  given  ratio  to  the  remainder 
FD. 

Because  the  ratio  of  AB  to  CD  is  given,  make  as  AB  to  CD, 
so  AG  to  CF:  therefore  the  ratio  of  AG  to  CF  is  given,  and  CF 
is  given,  wherefore  (2.  dat.)  AG  is 
given;    and   EA  is   given,  therefore     E         A  G         B 

the  whole  EG  is  given:  and  because     — 1 1 

as  AB  to  CD,  so  is  AG  to  CF,  and 

so  is  (19.  5.)  the  remainder  GB  to     C  F  D 

the  remainder  FD;  the  ratio  of  GB 


to  FD  is  given,  and  EG  is  given,  therefore  GB,  the  excess  of  the 
sum  EB  above  the  given  magnitude  EG,  has  a  given  ratio  to  the 
remainder  FD. 

PROP.  XXI.  C. 

If  two  magnitudes  have  a  given  ratio  to  one  ano- 
ther, if  a  given  magnitude  be  added  to  one  of  them, 
and  the  other  be  taken  from  a  given  magnitude;  the 
sum,  together  with  the  magnitude  to  which  the  remain- 
der has  a  given  ratio,  is  given;  and  the  remainder  is 
given  together  with  the  magnitude  to  which  the  sum 
has  a  given  ratio.* 

Let  the  two  magnitudes  AB,  CD  have  a  given  ratio  to  one  ano- 
ther; and  to  AB  let  the  given  magnitude  BE  be  added,  and  let 
CD  be  taken  from  the  given  magnitude  FD:  the  sum  AE  is  given, 
together  with  the  magnitude  to  which  the  remainder  FC  has  a 
given  ratio. 

Because  the  ratio  of  AB  to  CD  is  given,  make  as  AB  to  CD, 
so  GB  to  FD:  therefore  the  ratio  of  GB  to  FD  is  given,  and  FD 
is  given,  wherefore  GB  is  given  (2.  dat.); 
and   BE  is   given;    the   whole   GE   is     G     A  B         E 

therefore  given:  and  because  as  AB  to     1 1 

CD,  so  is  GB  to  FD,  and  so  is  (19.  5.) 

GA  to  FC;  the  ratio  of  GA  to  FC  is     F  C  D 

given:  and  AE  together  with  GA  is — 

*  See  Note. 


f 

'  Euclid's  data.  307 

given,  because  GE  is  givenj  therefore  the  sum  AE  together  with 
GA,  to  which  the  remainder  FC  has  a  given  ratio,  is  given.  The 
second  part  is  manifest  from  prop.  15. 

PROP.  XXII.  D. 

If  two  magnitudes  have  a  given  ratio  to  one  ano- 
ther, if  from  one  of  them  a  given  magnitude  be  taken, 
and  the  other  be  taken  from  a  given  magnitude  ;  each 
of  the  remainders  is  given,  together  with  the  magni- 
tude to  which  the  other  remainder  has  a  given  ratio.* 

Let  the  two  mai^nitudes  AB,  CD  have  a  given  ratio  to  one  ano- 
ther, and  from  AB  let  the  given  magnitude  AE  be  taken,  and  let 
CD  be  taken  from  the  given  magnitude  CF:  the  remainder  EB 
is  given,  together  with  the  magnitude  to  wliich  the  other  remain- 
der DF  has  a  given  ratio. 

Because  the  ratio  of  AB  to  CD  is  given,  make  as  AB  to  CD, 
so  AG  to  CF:  the  ratio  of  AG  to  CF  is  therefore  given,  and  CF 
is  given,  wherefore  (2.  dat.)  AG  is 
given^  and  AE  is  given,  and  there-     A  E  B  G 

fore  the  remainder  EG  is  given;     — — — 

and  because  as  AB  to  CD,  so  is 

AG  to  CF:  and  so  is  (19.  5.)  the     C  D 

remainder   BG   to  the  remainder     >— 


DF;  the  ratio  of  BG  to  DF  is  given:  and  EB  together  with  BG 
is  given,  because  EG  is  given:  therefore  the  remainder  EB  toge- 
ther with  BG,  to  which  DF  the  other  remainder  has  a  given  ratio, 
is  given.     The  second  part  is  plain  from  this  and  prop.  15. 

PROP.  XXIII.  20. 

If  from  two  given  magnitudes  there  be  taken  mag- 
nitudes which  have  a  given  ratio  to  one  another,  the 
remainders  shall  either  have  a  given  ratio  to  one  ano- 
ther, or  the  excess  of  one  of  them  above  a  given  mag- 
nitude shall  have  a  given  ratio  to  the  other."* 

Let  AB,  CD  be  two  given  magnitudes,  and  from  them  let  the 
magnitudes  AE,  CF,  which  have  a  given  ratio  to  one  another,  be 
taken;  the  remainders  EB,  FD  either  have  a  given  ratio  to  one 
another;  or  the  excess  of  one  of  them  above  a  given  magnitude 
has  a  given  ratio  to  the  other. 

Because   AB,   CD    are   each   of    A  E  B 

them  given,  the  ratio  of  AB  to  CD 1 

is  given:  and  if  this  ratio  be  the 

same  with  the  ratio  of  AE  to  CF,     C  F     D 

then  the  remainder  EB  has  (19.  5.) 

the  same  given  ratio  to  the  remainder  FD. 

*  See  Note. 


308  Euclid's  data. 

But  if  the  ratio  of  AB  to  CD  be  not  the  same  with  the  ratio  of 
AE  to  CF,  it  is  either  greater  than  it,  or,  by  inversion,  the  ratio 
of  CD  to  AB  is  greater  than  the  ratio  of  CF  to  AE:  first,  let  the 
ratio  of  AB  to  CD  be  greater  than  the  ratio  of  AE  to  CF;  and  as 
AE  to  CF,  so  make  AG  to  CD;  therefore  the  ratio  of  AG  to  CD 
is  given,  because  the  ratio  of  AE  to  CF  is  given;  and  CD  is  given, 
wherefore  (2.  dat.)  AG  is  given;  and  because  the  ratio  of  AB  to 
CD  is  greater  than  the  ratio  of  (AE  to 
CF,  that  is,  than  the  ratio  of)  AG  to     A  E  G     B 

CD;  AB  is  greater  (10.  5.)  than  AG:     1 j 

and  AB,  AG  are  given;  therefore  the 

remainder  BG  is  given:    and   because     X^  F       D 

as  AE  to  CF,  so  is  AG  to  CD,  and  so 


is  (19.  5.)  EG  to  FD;  the  ratio  of  EG  to  FE  is  given:  and  GB  is 
given;  therefore  EG,  the  excess  of  EB  above  a  given  magnitude 
GB,  has  a  given  ratio  to  FD*.  The  other  case  is  shown  in  the 
same  ^vay. 

PROP.  XXIV.  13. 

If  there  be  three  magnitudes,  the  first  of  which  has 
a  given  ratio  to  the  second,  and  the  excess  of  the  second 
above  a  given  magnitude  has  a  given  ratio  to  the  third; 
the  excess  of  the  first  above  a  given  magnitude  shall 
also  have  a  given  ratio  to  the  third.* 

Let  AB,  CD,  E,  be  the  three  magnitudes  of  which  AB  has  a 
given  ratio  to  CD;  and  the  excess  of  CD  above  a  given  magni- 
tude has  a  given  ratio  to  E:  the  excess  of  AB  above  a  given  mag- 
nitude has  a  given  ratio  to  E. 

Let  CF  be  the  given  magnitude,  the  excess  of  CD  above  which, 
viz.  FD  has  a  given  ratio  to  E:  and  because  the  ratio  of  AB  to 
CD  is  given,  as  AB  to  CD,  so  make  AG  to  A 
CF;  therefore  the  ratio  of  AG  to  CF  is  given; 
and  CF  is  given,  wherefore  (2.  dat.)  AG  is 
given:  and  because  as  AB  to  CD,  so  is  AG  ^ 
to  CF,  and  so  is  (19.  5.)  GB  to  FD;  the  ratio 
of  GB  to  FD  is  given.  And  the  ratio  of  FD 
to  E  is  given,  wherefore  (9.  dat.)  the  ratio  of 
GB  to  E  is  given,  and  AG  is  given;  therefore 
GB,  the  excess  of  AB  above  a  given  magni-  B 
tude  AG,  has  a  given  ratio  to  E. 

Cor.  1.  And  if  the  first  have  a  given  ratio  to  the  second,  and 
the  excess  of  the  first  above  a  given  magnitude  have  a  given  ratio 
to  the  third;  the  excess  of  the  second  above  a  given  magnitude 
shall  have  a  given  ratio  to  the  third.  For,  if  the  second  be  called 
the  first,  and  the  first  the  second,  this  corollary  will  be  the  same 
with  the  proposition. 

Cor.  2.  Also,  if  the  first  have  a  given  ratio  to  the  second,  and 
the  excess  of  the  third  above  a  given  magnitude  have  also  a  given 

*  See  Note. 


F— 


D 


E 


Euclid's  data.  309 

ratio  to  the  second,  the  same  excess  shall  have  a  given  ratio  to 
the  first;  as  is  evident  from  the  9th  dat. 

PROP.  XXV.  17. 

If  there  be  three  magnitudes,  the  excess  of  the  first 
whereof  above  a  given  magnitude  has  a  given  ratio  to 
the  second;  and  the  excess  of  the  third  above  a  given 
magnitude  has  a  given  ratio  to  the  same  second:  the 
first  shall  either  have  a  given  ratio  to  the  third,  or  the 
excess  of  one  of  them  above  a  given  magnitude  shall 
have  a  given  ratio  to  the  other. 

Let  AB,  C,  DE  be  three  magnitudes,  and  let  the  excesses  of 
each  of  the  two  AB,  DE  above  given  magnitudes*  have  given 
ratios  to  C;  AB,  DE  either  have  a  given  ratio  to  one  another, 
or  the  excess  of  one  of  them  above  a  given  magnitude  has  a  given 
ratio  to  the  other. 

Let  FB,  the  excess  of  AB  above  a  given  magnitude  AF,  have 
a  given  ratio  to  C;  and  let  GE,  the  excess      A 
of  DE  above  the  given  magnitude  DG,  have 
a  given  ratio  to  C;  and  because  FB,  GE  have 
each  of  them  a  given  ratio  to  C,  they  have      ■^' 
a  given  ratio  (9.  dat.)  to  one  another.    But  to 
FB,  GE  the  given  magnitudes  AF,  DG  are 
added;  therefore  (18.  dat.J  the  whole  magni- 
tudes AB,  DE  have  either  a  given  ratio  to  one 
another,  or  the  excess  of  one  of  them  above  a      B 
given  magnitude  has  a  given  ratio  to  the  other. 

PROP.  XXVL  18. 

If  there  be  three  magnitudes;  the  excesses  of  one  of 
which  above  given  magnitudes  have  given  ratios  to  the 
other  two  magnitudes;  these  two  shall  either  have  a 
given  ratio  to  one  another,  or  the  excess  of  one  of  them 
above  a  given  magnitude  shall  have  a  given  ratio  to  the 
other. 

Let  AB,  CD,  EF  be  three  magnitudes,  and  let  GD  the  excess 
of  one  of  them  CD  above  the  given  magnitude  CG  have  a  given 
ratio  to  AB;  and  also  let  KD  the  excess  of  the  same  CD  above  the 
given  magnitude  CK  have  a  given  ratio  to  EF:  either  AB  has  a 
given  ratio  to  EF,  or  the  excess  of  one  of  them  above  a  given  mag- 
nitude has  a  given  ratio  to  the  other. 

Because  GD  has  a  given  ratio  to  AB,  as  GD  to  AB,  so  make 
CG  to  HA;  therefore  the  ratio  of  CG  to  HA  is  given:  and  CG  is 
given,  wherefore  (2.  dat.)  HA  is  given;  and  because  as  GD  to  AB, 
so  is  CG  to  HA,  and  so  is  (12.  5.)  CD  to  HB;  the  ratio  of  CD  to 
HB  is  given:  also  because  KD  has  a  given  ratio  to  EF,  as  KD  to 


D 
G— 


310 


EUCLID  S   DATA. 


A— 


K-!- 


D 


EF,  so  make  CK  to  LE:  therefore  the  ratio     H 
of  CK  to  LE  is  given;  and   CK   is  given, 
wherefore  LE  (2.  dat.)  is  given:  and  because 
as  KD  to  EF,  so  is  CK  to  LE,  and  so  (12. 
5.)  is  CD  to  LF;  the  ratio  of  CD  to  LF  is  G-p         __ 

given:  but  the  ratio  of  CD  to  HB  is  given,  ir  I        -^   ~ 

wherefore  (9.  dat.)  the  ratio  of  HB  to  LF  is 
given:   and  from  HB,  LF  the   given  magni- 
tudes HA,  LE  being  taken,  the  remainders      3 
AB,  EF  shall  either  have  a  given  ratio  to  one 
another,  or  the  excess  of  one  of  them  above  a  given  magnitude 
has  a  given  ratio  to  the  other  (19.  dat.). 

Another  Demonstration, 

Let  AB,  C,  DE  be  three  magnitudes,  and  let  the  excesses  of 
one  of  them  C  above  given  magnitudes  have  given  ratios  to  AB 
and  DE:  either  AB,  DE  have  a  given  ratio  to  one  another,  or  the 
excess  of  one  of  them  above  a  given  magnitude  has  a  given  ratio 
to  the  other. 

Because  the  excess  of  C  above  a  given  magnitude  has  a  given 
ratio  to  AB;  therefore  (14.  dat.)  AB  together  with  a  given  mag- 
nitude has  a  given  ratio  to  C:  let  this  given  p 
magnitude  be  AF,  wherefore  FB  has  a  given 
ratio  to  C:  also  because  the  excess  of  C  above 
a  given  magnitude  has  a  given  ratio  to  DE;  ^ 
therefore  (14.  dat.)  DE  together  with  a  given 
magnitude  has  a  given  ratio  to  C:  let  this 
given  magnitude  be  DG,  wherefore  GE  has 
a  given  ratio  to  C:  and  FB  has  a  given  ratio 
to  C,  therefore  (9.  dat.)  the  ratio  of  FB  to  GE  is  given:  and  from 
FB,  GE  the  given  magnitudes  AF,  DG  being  taken,  the  remain- 
ders AB,  DE  either  have  a  given  ratio  to  one  another,  or  the 
excess  of  one  of  them  above  a  given  magnitude  has  a  given  ratio 
to  the  other  (l9.dat.). 


B 


G 

D 


PROP.  XXVII. 


19. 


If  there  be  three  magnitudes,  the  excess  of  the  first 
of  which  above  a  given  magnitude  has  a  given  ratio  to 
the  second;  and  the  excess  of  the  second  above  a  given 
magnitude  has  also  a  given  ratio  to  the  third;  the  ex- 
cess of  the  first  above  a  given  magnitude  shall  have  a 
given  ratio  to  the  third. 

Let  AB,  CD,  E  be  three  magnitudes,  the  excess  of  the  first  of 
which  AB  above  the  given  magnitude  AG,  viz.  GB,  has  a  given 
ratio  to  CD;  and  FD  the  excess  of  CD  above  the  given  magnitude 
CF,  has  a  given  ratio  to  E:  the  excess  of  AB  above  a  given  mag- 
nitude has  a  given  ratio  to  E. 

Because  the  ratio  of  GB  to  CD  is  given,  as  GB  to  CD,  so  make 


EUCLID  S    DATA. 


311 


G 


H— 


B 


F— 


D 


A  I 
E-r 

F— 


B 


D 


GH  to  CF:  therefore  the  ratio  of  GH  to  CF 
is  given;  afid  CF  is  given,  wherefore  (2.  dat.) 
GH  is  given:  and  AG  is  given,  wherefore  the 
whole  AH  is  given:  and  because  as  GB  to 
CD,  so  is  GH  to  CF,  and  so  is  (19.  5.)  the  re- 
mainder HB  to  the  remainder  FD;  the  ratio 
of  HB  to  FD  is  given:  and  the  ratio  of  FD  to 
E  is  given,  wherefore  (9.  dat.)  the  ratio  of  HB 
to  E  is  given:  and  AH  is  given;  therefore  HB, 
the  excess  of  AB  above  a  given  magnitude  AH,  has  a  given  ratio 
toE. 

^  "  Otherwise^ 

Let  AB,  C,  D,  be  three  magnitudes,  the  excess  EB  of  the  first 
of  which  AB  above  the  given  magnitude  AE  has  a  given  ratio  to 
C,  and  the  excess  of  C  above  a  given  magni- 
tude has  a  given  ratio  to  D:  the  excess  of  AB 
above  a  given  magnitude  has  a  given  ratio 
to  D. 

Because  EB  has  a  given  ratio  to  C,  and  the 
excess  of  C  above  a  given  magnitude  has  a 
given  ratio  to  D;  therefore  (24.  dat.)  the  ex- 
cess of  EB  above  a  given  magnitude  has  a 
given  ratio  to  D:  let  this  given  magnitude 
be  EF;  therefore  FB,  the  excess  of  EB  above  EF,  has  a  given 
ratio  to  D:  and  AF  is  given,  because  AE,  EF  are  given:  therefore 
FB,  the  excess  of  AB  above  a  given  magnitude  AF,  has  a  given 
ratio  to  D." 

PROP.  XXVni.  25. 

If  two  lines  given  in  position  cut  one  another,  the 
point  or  points  in  which  they  cut  one  another  are  given.* 

Let  two  lines  AB,  CD  given  in  position  cut  one  another  in  the 
point  E;  the  point  E  is  given.  C 

Because  the  lines  AB,  CD  are 
given  in  position,  they  have  always 
the  same  situation  (4.  def.),  and 
therefore  the  point,  or  points,  in 
which  they  cut  one  another,  have  al- 
ways the  same  situation:  and  because 
the  lines  AB,  CD  can  be  found  (4. 
def.),  the  point,  or  points,  in  which 
they  cut  one  another,  are  likewise 
found;  and'therefore  are  given  in  po- 
sition (4.  def.) 

PROP.  XXIX.  26. 

If  the  extremities  of  a  straight  hne  be  given  in  posi- 
tion; the  straight  hne  is  given  in  position  and  magni- 
tude. 

Because  the  extremities  of  the  straight  line  are  given,  they  can 


A— 


*  See  Note. 


312  Euclid's  DATA.  *^       --,  jt*'***    - 

be  found  (4.  def.):  let  these  be  the  points  A,  B,  between  which 

a  straight  line  AB  can  be  drawn  (l.pos-  *__ -d 

tulate)j  this  lias  an  invariable  position, 

because  between  two  given  points  there  can  be  drawn  but  one 
straight  line:  and  when  the  straight  line  AB  is  drawn,  its  magni- 
tude is  at  the  same  time  exhibited,  or  given:  therefore  the  straight 
line  AB  is  given  in  position  and  magnitude. 

PROP.  XXX.  27. 

If  one  of  the  extremities  of  a  straight  line  given  in 
position  and  magnitude  be  given :  the  other  extrejjaity 
shall  also  be  given. 

Let  the  point  A  be  given,  to  wit,  one  of  the  extremities  of  a 
straight  line  given  in  magnitude,  and  which  lies  in  the  straight 
line  AC  given  in  position^  the  other  extremity  is  also  given. 

Because  the  straight  line  is  given  in  magnitude,  one  equal  to  it 
can  be  found  (l.  def.);  let  this  be  the  straight  line  D:  from  the 
greater  straight  line  AC  cut  off  AB  equal      »  R        C 

to  the  lesser  D:  therefore  the  other  ex-     

tremity  B  of  the  straight   line  AB    is 
found:  and  the  point  B  has  always  the     D 
same  situation;  because  any  other  point 


in  AC,  upon  the  same  side  of  A,  cuts  off  between  it  and  the  point 
A  a  greater  or  less  straight  line  than  AB,  that  is,  than  D;  therefore 
the  point  B  is  given  (4.  def.):  and  it  is  plain  another  such  point  can 
be  found  in  AC,  produced  upon  the  other  side  of  the  point  A. 

PROP.  XXXI.  28. 

If  a  straight  line  be  dravi^n  through  a  given  point  pa- 
rallel to  a  straight  line  given  in  position;  that  straight 
line  is  given  in  position. 

Let  A  be  a  given  point,  and  BC  a  straight  line  given  in  position; 
the  straight  line  drawn  through  a  parallel  to  BC  is  given  in  po- 
sition. 

Through  A  draw  (31.  1.)  the  straight     j^  A         E 

line  DAE  parallel  to  BC;  the  straight ■ 

line  DAE  has  always  the  same  position,  ' 

because  no  other  straight   line   can  be     B  C 

drawn  through  A  parallel  to  BC:  there- 


fore the  straight  line  DAE,  which  has  been  found,  is  given  (4.  def.) 
in  position. 

PROP.  XXXn.  29. 

If  a  straight  line  be  drawn  to  a  given  point  in  a 
straight  line  given  in  position,  and  makes  a  given  angle 
with  it;  that  straight  line  is  given  in  position. 

Let  AB  be  a  straight  line  given  in  position,  and  C  a  given  point 


^        "      Euclid's  DATA.  313 

ill  it;  the  straight  line  drawn  to  C, 
which  makes  a  given  angle  with  CB, 
is  given  in  position. 

Because  the  angle  is  given,  one 
equal  to  it  can  be  found  (l.  def.);  let 
this  be  the  angle  at  D:  at  the  given      a*~~"  q  B 

point  C,  in  the  given  straight  line 
AB,  make  (23.  I.)  the  angle  ECB 
equal  to  the  angle  at  D:  therefore 
the  straight  line  EC  has  always  the 
same  situation,  because  any  other 
straight  line  FC,  drawn  to  the  point  ^ 

C,  makes  with  CB  a  greater  or  less  angle  than  the  angle  ECB,  or 
the  angle  at  D:  therefore  the  straight  line  EC,  which  has  been 
found,  is  given  in  position. 

It  is  to  be  observed,  that  there  are  two  straight  lines  EC,  GC 
upon  one. side  of  AB  that  make  equal  angles  with  it,  and  which 
make  equal  angles  with  it  when  produced  to  the  other  side. 

PROP.  XXXIII.  30. 

If  a  straight  line  be  drawn  from  a  given  point  to  a 
straight  hne  given  in  position,  and  makes  a  given  angle 
with  it;  that  straight  line  is  given  in  position. 

From  the  given  point  A,  let  the  straight  line  AD  be  drawn  to 
the  straight  line  BC  given  in  position,  and  make  with  it  a  given 
angle  ADC;  AD  is  given  in  position.  E  A  F 

Through  the  point  A,  draw  (31.  1.)  the 
straight  line  EAF  parallel  to  BC;  and 
because  through  the  given  point  A,  the 
straight  line  EAF  is  drawn  parallel  to  BC, 
which  is  given  in  position,  EAF  is  there-     ^  DC 

fore  given  in  position  (31.  dat.):  and  because  the  straight  line  AD 
meets  the  parallels,  BC,  EF,  the  angle  EAD  is  equal  (29.  1.)  to 
the  angle  ADC;  and  ADC  is  given,  wherefore  also  the  angle  EAD 
is  given:  therefore,  because  the  straight  line  DA  is  drawn  to  the 
given  point  A  in  the  straight  line  EF  given  in  position,  and 
makes  with  it  a  given  angle  EAD,  AD  is  given  (32.  dat.)  in 
position. 

PROP.  XXXIV.  31. 

If  from  a  given  point  to  a  straight  line  given  in  posi- 
tion, a  straight  line  be  drawn  which  is  given  in  magni- 
tude; the  same  is  also  given  in  position."* 

Let  A  be  a  given  point,  and  BC  a  straight  line  given  in  posi- 
tion; a  straight  line  given  in  magnitude  drawn  from  the  point  A 
to  BC  is  given  in  position. 

Because  the  straight  line  is  given  in  magnitude,  one  equal  to 

^  See  Note. 
2  R 


314  Euclid's  data.    •         '•'' 

it  can  be  found  (l.  clef.);  let  this  be  the  straight  line  D:  from  the 
point  A  draw  AE  perpendicular  to  BC,*  and  be-  A 

cause  AE  is  the  shortest  of  all  the  straight  lines 
which  can  be  drawn  from  the  point  A  to  BC,  the 
straight  line  D,  to  which   one  equal  is  to  be 

drawn  from  the  point  A  to  BC,  cannot  be  less     

than  AE.     If  therefore  D  be  equal  to  AE,  AE     g  £  C 

is  the  straight  line  given  in  magnitude,  drawn     j^ 

from  the  given  point  A  to  BC:  and  it  is  evident 
that  AE  is  given  in  position,  (33.  dat.),  because  it  is  drawn  from 
the  given  point  A.  to  BC,  which  is  given  in  position,  and  makes 
with  BC  the  given  angle  AEC. 

But  if  the  straight  line  D  be  not  equal  to  AE,  it  must  be  greater 
than  it:  produce  AE,  and  make  AF  equal  to  D;  and  from  the 
centre  A,  at  the  distance  AF,  describe  the  circle  GFH,  and  join 
AG,  AH:  because  the  circle  GFH  is  given  in  position  (6.  def.), 
and  the  straight  line  BC  is  also  given  in  position;  therefore  their 
intersection  G  is  given  (28.  dat.);  A 

and  the  point  A  is  given;  where- 
fore AG  is  given  in  position  (29. 
dat.),  that  is,  the  straight  line 
AG  given  in  magnitude,  (for  it 
is  equal  to  D)  and  drawn  from 
the  given  point  A  to  the  straight 
line  BC  given  in  position,  is  also 
given  in  position:  and  in  like  manner  AH  is  given  in  position: 
therefore  in  this  case  there  are  two  straight  lines  AG,  AH  of  the 
same  given  magnitude,  which  can  be  drawn  from  a  given  point  A 
to  a  straight  line  BC  given  in  position. 

PROP.  XXXV.  32. 

If  a  straight  line  be  drawn  between  two  parallel 
straight  lines  given  in  position,  and  makes  given  angles 
with  them,  the  straight  line  is  given  in  magnitude. 

Let  the  straight  line  EF  be  drawn  between  the  parallels  AB, 
CD,  which  are  given  in  position,  and  make  the  given  angles  BEF, 
EFD:  EF  is  given  in  magnitude. 

In  CD  take  the  given  point  G,  and  through  G  draw  (31.  1.)  GH 
parallel  to  EF:  and  because  CD  meets  the  parallels  GH,  EF,  the 
angle  EFD  is  equal  (29.  1.)  to  the  angle      .  EH  B 

HGD:  and  EFD  is  a  given  angle; 
wherefore  the  angle  FIGD  is  given;  and 
because  HG  is  drawn  to  the  given  point 
G,  in  the  straight  line  CD,  given  in  po- 
sition, and  makes  a  given  angle  HGD:  p  ^  "A  ~^ 
the  straight  line  HG  is  given  in  posi- 
tion (32.  dat.):  and  AB  is  given  in  position:  therefore  the  point 
H  is  given  (28.  dat.),  and  the  point  G  is  also  given,  wherefore  GH 
is  given  in  magnitude  (29.  dat.)  and  EF  is  equal  to  it,  therefore 
EF  is  given  in  magnitude. 


Euclid's  data.  315 


PROP.  XXXVI.  33. 

If  a  straight  line  given  in  magnitude  be  drawn  be- 
tween two  parallel  straight  lines  given  in  position,  it 
shall  make  given  angles  with  the  parallels.* 

Let  the  straight  line  EF  given  in  magnitude  be  drawn  between 
the  parallel  straight  lines  AB,  CD,  which  are      .  F     R      n 

given  in  position:  the  angles  AEF,  EFC  shall 
be  given. 

Because  EF  is  given  in  magnitude,  a  straight 
line  equal  to  it  can  be  found  (1.  def.):  let  this 
be  G:  in  AB  take  a  given  point  H,  and  from 


it  draw  (12.   1.)  HK  perpendicular  to  CD-,     C  F    K      D 

therefore  the  straight  line  G,  that  is,  EF,  can-  G 

not  be  less  than  HK:  and  if  G  be  equal  to  HK,  EF  also  is  equal 
to  it:  wherefore  EF  is  at  right  angles  to  CD:  for  if  it  be  not,  EF 
would  be  greater  than  HK,  which  is  absurd.  Therefore  the  angle 
EFD  is  a  right,  and  consequently  a  given  angle. 

But  if  the  straight  line  G  be  not  equal  to  HK,  it  must  be  greater 
than  it:  produce  HK,  and  take  HL,  equal  to  G,  and  from  the  centre 
H,  at  the  distance  HL,  describe  the  circle  MLN,  and  join  HM, 
HN:  and  because  the  circle  (6.  def.)  MLN,  and  the  straight  line 
CD,  are  given  in  position,  the  points  M,  N  are  (28.  dat.)  given: 
and    the   point   H   is    given, 

wherefore   the  straight  lines     A  E  H  B 

HM,  HN,  are  given  in  posi- 
tion (29.  dat.)  and  CD  is  given 
in  position:  therefore  the  an- 
gles HMN,  HNM,  are  given 
in  position  (A.  def.):  of  the 
straight  lines  HM,  HN,  let 

HN  be  that  which  is  not  pa-  G 

rallel  to  EF,  for  EF  cannot  be 

parallel  to  both  of  them;  and  draw  EO  parallel  to  HN:  EO  there- 
fore is  equal  (34.  1.)  to  HN,  that  is  to  G;  and  EF  is  equal  to  G, 
wherefore  EO  is  equal  to  EF,  and  the  angle  EFO  to  the  angle 
EOF,  that  is,  (29.  1.),  to  the  given  angle  HNM;  and  because  the 
angle  HNM,  which  is  equal  to  the  angle  EFO,  or  EFD,  has  been 
found:  therefore  the  angle  EFD,  that  is,  the  angle  AEF,  is  given 
in  magnitude  (1.  def.);  and  consequently  the  angle  EFC. 

PROP.  XXXVH.  E. 

If  a  straight  line  given  in  magnitude  be  drawn  from 
a  point  to  a  straight  line  given  in  position,  in  a  given 
angle;  the  straight  line  drawn  through  that  point  pa- 
rallel to  the  straight  line  given  in  position,  is  given  in 
position.* 

*  See  Note. 


316  Euclid's  data. 

Let  the  straight  line  AD  given  in  magnitude  be  drawn  from 
the  point  A  to  the  straight  line  BC  given    E         A     H  F 

in  position,  in  the  given  angle  ADC:  the     — 
straight  line  EAF  drawn  through  A  pa- 
rallel to  BC  is  given  in  position. 

In  BC  take  a  given  point  G,  and  draw 

GH  parallel  to  AD:  and  because  HG  is     B     D     G  C 

drawn  to  a  given  point  G  in  the  straight  line  BC  given  in  posi- 
tion, in  a  given  angle  HGC,  for  it  is  equal  (29.  1.)  to  the  given 
angle  ADC|  HG  is  given  in  position  (32.  dat.)^  but  it  is  given 
also  in  magnitude,  because  it  is  equal  to  (34.  1.)  AD  which  is 
given  in  magnitude;  therefore  because  G,  one  of  the  extremities 
of  the  straight  line  GH,  given  in  position  and  magnitude  is  given, 
the  other  extremity  H  is  given  (30.  dat.);  and  the  straight  line 
EAF,  which  is  drawn  through  the  given  point  H  parallel  to  BC 
given  in  position,  is  therefore  given  (31.  dat.)  in  position. 

PROP.  XXXVm.  34. 

If  a  straight  line  be  drawn  from  a  given  point  to  two 
parallel  straight  lines  given  in  position,  the  ratio  of  the 
segments  between  the  given  point  and  the  parallels 
shall  be  given. 

Let  the  straight  line  EFG  be  drawn  from  the  given  point  E  to 
the  parallels  AB,  CD;  the  ratio  of  EF  to  EG  is  given. 

From  the  point  E  draw  EHK  perpendicular  to  CD;  and  be- 
cause from  a  given  point  E  the  straight  line  EK  is  drawn  to  CD 
which  is  given  in  position,  in  a  given  angle  EKC;  EK  is  given  in 

E 


B 


F    H  B 


CGK  DC  KGD 

position  (33.  dat.);  and  AB,  CD  are  given  in  position:  therefore 
(28.  dat.)  the  points  H,  K  are  given;  and  the  point  E  is  given; 
wherefore  (29.  dat.)  EH,  EK  are  given  in  magnitude,  and  the 
ratio  (1.  dat.)  of  them  is  therefore  given.  But  as  EH  to  EK,  so 
is  EF  to  EG,  because  AB,  CD  are  parallels;  therefore  the  ratio 
of  EF  to  EG  is  given. 

PROP.  XXXIX.  35,  36. 

If  the  ratio  of  the  segments  of  a  straight  line  be- 
tween a  given  point  in  it  and  two  parallel  straight  lines 
be  given,  if  one  of  the  parallels  be  given  in  position,  the 
other  is  also  given  in  position. 


Euclid's  data. 


317 


From  the  given  point  A,  let  the  straight  line  AED  be  drawn  to 
the  two  parallel  straight  lines  FG,  BC,  and  let  the  ratio  of  the 
segments  AE,  AD  be  given;  if  one  of  the  parallels  BC  be  given 
in  position,  the  other  FG  is  also  given  in  position. 

From  the  point  A,  draw  AH  perpendicular  to  BC,  and  let  it 
meet  FG  in  K:  and  because  AH  is  drawn  from  the  given  point  A 
to  the  straight  line  BC  given  in  position,  and  makes  a  given  angle 

A 


E     K 


G 


B 


D 


H 


B 


H 


D 


AHD;  AH  is  given  (33.  dat.)  in  posi- 
tion; and  BC  is  likewise  given  in  posi- 
tion: therefore  the  point  H  is  given 
(28.  dat.):  the  point  A  is  also  given; 
wherefore  AH  is  given  in  magnitude 
(29.  dat.);  and  because  FG,  BC  are  pa- 
rallels, as  AE  to  AD,  so  is  AK  to  AH; 
and  the  ratio  of  AE  to  AD  is  given,      F         E  KG 

wherefore  the  ratio  of  AK  to  AH  is  given;  but  AH  is  given  in  mag- 
nitude, therefore  (2.  dat.)  AK  is  given  in  magnitude;  and  it  is 
also  given  in  position,  and  the  point  A  is  given;  wherefore  (30. 
dat.)  the  point  K  is  given.  And  because  the  straight  line  FG  is 
drawn  through  the  given  point  K  parallel  to  BC  which  is  given 
in  position,  therefore  (31.  dat.)  FG  is  given  in  position. 


PROP.  XL. 


37,  38. 


If  the  ratio  of  the  segments  of  a  straight  Hne  into 
which  it  is  cut  by  three  parallel  straight  lines,  be  given; 
if  two  of  the  parallels  are  given  in  position,  the  third 
is  also  given  in  position.* 

Let  AB,  CD,  HK  be  three  parallel  straight  lines,  of  which  AB, 
CD  are  given  in  position;  and  let  the  ratio  of  the  segments  GE, 
GF  into  which  the  straight  line  GEF  is  cut  by  the  three  paral- 
lels, be  given;  the  third  parallel  HK  is  given  in  position. 

In  AB  take  a  given  point  L,  and  draw  LM  perpendicular  to 
CD,  meeting  HK  in  N;  because  LM  is  drawn  from  the  given 
point  L  to  CD  which  is  given  in  position,  and  makes  a  given  an- 
gle LMD;  LM  is  given  in  position  (33.dat.);  and  CD  is  given  in 
position,  wherefore  the  point  M  is  given  (28.  dat.);  and  the  point 
L  is  given;  LM  is  therefore  given  in  magnitude  (29.  dat.):  and 
because  the  ratio  of  GE  to  GF  is  given,  and  as  GE  to  GF,  so  is 


See  Note. 


318 


H 


G     N 


EUCLID  S  DATA. 

K  A 


H 


B 


G    N 


K 


M 


D 


NL  to  NM;  the  ratio  of  NL  to  NM  is  given;  and  therefore  (cor. 
6.  or  7.  dat.)  the  ratio  of  ML  to  LN  is  given;  but  LM  is  given  in 
magnitude  (cor.  6.  or  7.  dat.),  wherefore  (2.  dat.)  LN  is  given  in 
magnitude;  and  it  is  also  given  in  position,  and  the  point  L  is 
given,  wherefore  (30.  dat.)  the  point  N  is  given;  and  because  the 
straight  line  HK  is  drawn  through  the  given  point  N  parallel  to 
CD  which  is  given  in  position,  therefore  HK  is  given  in  position 
(31.dat.). 

PROP.  XLI.  F. 

If  a  straight  line  meets  three  parallel  straight  lines 
which  are  given  in  position,  the  segments  into  which 
they  cut  it  have  a  given  ratio. 

Let  the  parallel  straight  lines  AB,  CD,  EF,  given  in  position, 
be  cut  by  the  straight  line  GHK;  the  ratio  of  GH  to  HK  is  given. 

In  AB  take  a  given  point  L,  and       A  G      L  B 

draw  LM  perpendicular  to  CD,  meet-        -— 
ing  EF  in  N;  therefore  (33.  dat.)  LM 
is  given  in  position;  and  CD,  EF  are      ^ 
given  in  position,  wherefore  the  points 
M,  N  are  given;  and  the  point  L  is 
given;  therefore  (29.  dat.)  the  straight      . 
lines  LM,  MN  are  given  in  magni-      E       K  N  F 

tude;  and  the  ratio  of  LM  to  MN  is  therefore  given  (1.  dat.):  but 
as  LM  to  MN,  so  is  GH  to  HK;  wherefore  the  ratio  of  GH  to 
HK  is  given. 


M 


D 


PROP.  XLH. 


39. 


If  each  of  the  sides  of  a  triangle  be  given  in  magni- 
tude, the  triangle  is  given  in  species. 

Let  each  of  the  sides  of  the  triangle  ABC  be  given  in  magni- 
tude, the  triangle  ABC  is  given  in  species. 

Make  a  triangle  (22.  1.)  A  D 
DEF,  the  sides  of  which  are 
equal,  each  to  each,  to  the 
given  straight  lines  AB,  BC, 
CA,  which  can  be  done;  be- 
cause any  two  of  them  must     ,    , 

be  greater  than  the  third;  and      B  C  E  r 

let  DE  be  equal  to  AB,  EF  to  BC,  and  FD  to  CA;  and  because 
the  two  sides  ED,  DF  are  equal  to  the  two  BA,  AC,  each  to  each, 
and  the  base  EF  equal  to  the  base  BC;  the  angle  EDF  is  equal 


Euclid's  data.  319 

(8.  1.)  to  the  angle  BAG;  therefore,  because  the  angle  EDF  which 
is  equal  to  the  angle  BAG,  has  been  found,  the  angle  BAG  is 
given  (l.  def.);  in  like  manner  the  angles  at  B,  G  are  given.  And 
because  the  sides  AB,  BG,  GA  are  given,  their  ratios  to  one  an- 
other are  given  (1.  dat.),  therefore  the  triangle  ABG  is  given  (3. 
def)  in  species. 

PROP.  XLIII.  40. 

If  each  of  the  angles  of  a  triangle  be  given  in  mag- 
nitude, the  triangle  is  given  in  species. 

Let  each  of  the  angles  of  the  triangle  ABG  be  given  in  mag- 
nitude, the  triangle  ABG  is  given  in 
species.  A 

Take  a  straight  line  DE  given  in 
position  and  magnitude,  and  at  the 
points  D,  E  make  (23.  1.)  the  angle 
EDF  equal  to  the  angle  BAG;  and 
the  angle  DEF  equal  to  ABG;  there- 
fore the  other  angles  EFD,  BGA  are  ^  C  E  F 
equal,  and  each  of  the  angles  at  the  points  A,  B,  G  is  given; 
wherefore  each  of  those  at  the  points  D,  E,  F  is  given:  and  be- 
cause the  straight  line  FD  is  drawn  to  the  given  point  D  in  DE, 
which  is  given  in  position,  making  the  given  angle  EDF;  there- 
fore DF  is  given  in  position  (32.  dat.).  In  like  manner  EF  also 
is  given  in  position;  wherefore  the  point  F  is  given:  and  the 
points  D,  E  are  given;  therefore  each  of  the  straight  lines  DE, 
EF,  FD  is  given  (29.  dat.)  in  magnitude;  wherefore  the  triano-le 
DEF  is  given  in  species  (42.  dat.);  and  it  is  similar  (4.  6.  1.  def. 
6.)  to  the  triangle  ABG:  which  is  therefore  given  in  species. 

PROP.  XLIV.  41. 

If  one  of  the  angles  of  a  triangle  be  given,  and  if  the 
sides  about  it  have  a  given  ratio  to  one  another;  the  tri- 
angle is  given  in  species. 

Let  the  triangle  ABG  have  one  of  its  angles  BAG  given,  and 
let  the  sides  BA,  AG  about  it  have  a  given  ratio  to  one  another; 
the  triangle  ABC  is  given  in  species. 

Take  a  straight  line  DE  given  in  position  and  magnitude,  and 
at  the  point  D,  in  the  given  straight  line  DE,  make  the  angle 
EDF  equal  to  the  given  angle  BAG;  wherefore  the  angle  EDF 
is  given;  and  because  the  straight  line  FD  is  drawn  to  the  given 
point  D  in  ED  which  is  given  in  position,  making  the  given 
angle  EDF;  therefore  FD  is  given  in  A 

position  (32.  dat.).     And  because  the  .  j. 

ratio  of  BA  to  AC  is  given,  make  the 
ratio  of  ED  to  DF  the  same  with  it, 
and  join  EF;  and  because  the  ratio  of 
ED  to  DF  is  given,  and  ED  is  given, 

therefore  (2.  dat.)  DF  is  given  in  mag-     B  C       E 

nitude:  and  it  is  given  also  in  position,  and  the  point  D  is  given. 


320 


EUCLID  S    DATA. 


wherefore  the  point  F  is  given  (30.  dat.);  and  the  points  D,  E 
are  given,  wherefore  DE,  EF,  FD  are  given  (29.  dat.)  in  mag- 
nitudej  and  the  triangle  DEF  is  therefore  given  (^42.  dat.)  in  spe- 
cies; and  because  the  triangles  ABC,  DEF  have  one  angle  BAG 
equal  to  one  angle  EDF,  and  the  sides  about  these  angles  pro- 
portionals; the  triangles  are  (6.  6.)  similar;  but  the  triangle  DEF 
is  given  in  species,  and  therefore  also  the  triangle  ABC. 


PROP.  XLV. 


42. 


DEF 


If  the  sides  of  a  triangle  have  to  one  another  given 
ratios,  the  triangle  is  given  in  species. 

Let  the  sides  of  the  triangle  ABC  have  given  ratios  to  one  an- 
other, the  triangle  ABC  is  given  in  species. 

Take  a  straight  line  D  given  in  magnitude;  and  because  the 
ratio  of  AB  to  EC  is  given,  make  the  ratio  of  D  to  E  the  same 
with  it;  and  D  is  given,  therefore  (2.  dat.)  E  is  given.     And  be- 
cause the  ratio  of  BC  to  CA  is  given,  to  this  make  the  ratio  of  E 
to  F  the  same;  and  E  is  given,  and  therefore  (2.  dat.)  F;  and  be- 
cause as  AB  to  BC,  so  is  D  to  E;  by  composition  AB  and  BC 
together  are  to  BC,  as  D  and  E  to 
F;  but   as  BC  to  CA,   so  is  E  to 
F;  therefore,  ex  sequali^  (22.  5.)  as 
AB  and  BC  are  to  CA,  so  are  D 
and  E  to  F,  and  AB  and  BC  are 
greater  (20.  1.)  than  CA;  therefore 
D  and  E  are  greater  (A.  5.)  than 
F.     In  the  same  manner  any  two 
of  the  three  D,  E,   F  are  greater 
than  the  third.     Make  (22.  1.)  the 
triangle    GHK    whose    sides    are 
equal  to  D,  E,  F,  so  that  GH  be 
equal  to  D,  HK  to  E,  and  KG  to 
F;  and  because  D,  E,  F  are  each     H         K 
of  them  given,  therefore  GH,  HK,  KG  are  each  of  them  given  in 
magnitude;  therefore  the  triangle  GHK  is  given  (42.  dat.)  in  spe- 
cies; but  as  AB  to  BC,  so  is  (D  to  E,  that  is)  GH  to  HK;  and  as 
BC  to  CA,  so  is  (E  to  F,  that  is)  HK  to  KG;  therefore,  ex  sequali, 
as  AD  to  AC,  so  is  GH  to  GK.     Wherefore  (5.  6.)  the  triangle 
ABC  is  equiangular  and  similar  to  the  triangle  GHK;  and  the 
triangle  GHK   is  given  in  species;  therefore    also  the   triangle 
ABC  is  given  in  species. 

Cor.  If  a  triangle  is  required  to  be  made;  the  sides  of  which 
shall  have  the  same  ratios  which  three  given  straight  lines  D,  E, 
F  have  to  one  another;  it  is  necessary  that  every  two  of  them  be 
greater  than  the  third. 

PROP.  XLVI.  43. 

If  the  sides  of  a  right  angled  triangle  about  one  of 
the  acute  angles  have  a  given  ratio  to  one  another;  the 
triangle  is  given  in  species. 


Euclid's  data.  321 

Let  the  sides  AB,  BC  about  the  acute  angle  ABC  of  the  tri- 
angle ABC,  which  has  a  right  angle  at  A,  have  a  given  ratio  to 
one  another;  the  triangle  ABC  is  given  in  species. 

Take  a  straight  line  DE  given  in  position  and  magnitude;  and 
because  the  ratio  of  AB  to  BC  is  given,  make  as  AB  to  BC,  so 
DE  to  EF;  and  because  DE  has  a  given  ratio  to  EF,  and  DE  is 
given,  therefore  (2.  dat.)  EF  is  given;  and  because  as  AB  to  BC, 
so  is  DE  to  EF;  and  AB  is  less  (19.  1.)  than  BC,  therefore  DE  is 
less  (A.  5.)  than  EF.     From  the  point  D  draw  DG  at  right  angles 
to  DE,  and  from  the  centre  E,  at 
the  distance  EF,  describe  a  circle 
which    shall    meet    DG    in    two 
points;  let  G  be  either  of  them, 
and  join  EG;   therefore   the  cir- 
cumference of  the  circle  is  given 
(6.    def.)    in    position;    and    the 
straight   line    DG    is    given   (32. 
dat.)  in  position,  because  it  is  drawn  to  the  given  point  D  in  DE 
given  in  position,  in  a  given  angle;  therefore  (28.  dat.)  the  point 
G  is  given;  and  the  points  D,  E  are  given:  wherefore  DE,  EG, 
GD  are  given  (29.  dat.)  in  magnitude,  and  the  triangle  DEG  in 
species  (42.  dat.).     And  because  the  triangles  ABC,  DEG  have 
the  angle  BAC,  equal  to  the  angle  EDG,  and  the  sides  about  the 
angles  ABC,  DEG  proportionals,  and  each  of  the  other  angles 
BCA,  EGD  less  than  a  right  angle;  the  triangle  ABC  is  equi- 
angular (7.  6.)  and  similar  to  the  triangle  DEG:  but  DEG  is  given 
in  species;  therefore  the  triangle  ABC  is  given  in  species:  and,  in 
the  same  manner,  the  triangle  made  by  drawing  a  straight  line 
from  E  to  the  other  point  in  which  the  circle  meets  DG  is  given 
in  species. 

PROP.  XLVII.  44. 

If  a  triangle  has  one  of  its  angles  which  is  not  a  right 
angle  given,  and  if  the  sides  about  another  angle  have 
a  given  ratio  to  one  another;  the  triangle  is  given  in 
species. 

Let  the  triangle  ABC  have  one  of  its  angles  ABC  a  given  but 
not  a  right  angle,  and  let  the  sides  BA,  AC  about  another  angle 
BAC  have  a  given  ratio  to  one  another;  the  triangle  ABC  is  given 
in  species. 

First,  let  the  given  ratio  be  the  ratio  of 
equality,  that  is,  let  the  sides  BA,  AC  and  con- 
sequently the  angles  ABC,  ACB  be  equal;  and 
because  the  angle  ABC  is  given,  the  angle  ACB, 
and  also  the  remaining  (32.  1.)  angle  BAC  is 
given;  therefore  the  triangle  ABC  is  given  (43. 
dat.)  in  species;  and  it  is  evident  that  in  this 
case  the  given  angle  ABC  must  be  acute. 

Next,  let  the  given  ratio  be  the  ratio  of  a  less  to  a  greater, 
that  is,  let  the  side  AB  adjacent  to  the  given  angle  be  less  than 
2   S 


322 


EUCLID  S    DATA. 


A 


B 


D 


because  BA  is  less 


the  side  AC;  take  a  straight  line  DE  given  in  position  and  mag- 
nitude, and  make  the  angle  DEF  equal  to  the  given  angle  ABC; 
therefore  EF  is  given  (32.  dat.)  in  position;  and  because  the 
ratio  of  BA  to  AC  is  given,  as  BA  to 
AC,  so  make  ED  to  DG;  and  because 
the  ratio  of  ED  to  DG  is  given,  and 
ED  is  given,  the  straight  line  DG  is 
given  (2.  dat.),  and  BA  is  less  than 
AC,  therefore  ED  is  less  (A.  5.)  than 
DG.  From  the  centre  D  at  the  dis- 
tance DG  describe  the  circle  GF  meet- 
ing EF  in  F,  and  join  DF;  and  be- 
cause the  circle  is  given  (6.  def,)  in 
position,  as  also  the  straight  line  EF, 
the  point  F  is  given  (28.  dat.);  and 
the  points  D,  E  are  given;  w^herefore 


the  straight  lines  DE,  EF,  FD  are 
given  (29.  dat.)  in  magnitude,  and  the 
triangle  DEF  in  species  (42.  dat.). 
than  AC,  the  angle  ACB  is  less  (18.  l.)than  the  angle  ABC, 
and  therefore  ABC  is  less  (l.  7.  1.)  than  a  right  angle.  In  the 
same  manner,  because  ED  is  less  than  DG  or  DF,  the  angle 
DFE  is  less  than  a  right  angle:  and  because  the  triangles  ABC, 
DEF  have  the  angle  ABC  equal  to  the  angle  DEF,  and  the  sides 
about  the  angles  BAC,  EDF  proportionals,  and  each  of  the  other 
angles  ACB,  DFE  less  than  a  right  angle;  the  triangles  ABC, 
DEF  are  (7.  6.)  similar,  and  DEF  is  given  in  species,  wherefore 
the  triangle  ABC  is  also  given  in  species. 

Thirdly,  let  the  given  ratio  be  the  ratio  of  a  greater  to  a  less, 
that  is,  let  the  side  AB  adjacent  to  the  given  angle  be  greater 
than  AC;  and,  as  in  the  last  case,  take  a 
straight  line  DE    given  in   position   and  A 

magnitude,  and  make  the  angle  DEF  equal 
to  the  given  angle  ABC;  therefore  EF  is 
given  (32.  dat.)  in  position:  also  draw 
DG  perpendicular  to  EF;  therefore  if  the 
ratio  of  BA  to  AC  be  the  same  with  the 
ratio  ofJEDito  the  perpendicular  DG, 
the  triangles  ABCj  DEG  are  similar 
(7.  6.),  because  the  angles  ABC,  DEG 
are  equal,  and  DGE  is  a  right  angle: 
therefore  the  angle  ACB  is  a  right  angle, 
and  the  triangle  ABC  is  given  in  (43.  dat.) 
species. 

But  if,  in  this  last  case,  the  given  ratio  of  BA  to  AC  be  not 
the  same  with  the  ratio  of  ED  to  DG,  that  is,  with  the  ratio 
of  BA  to  the  perpendicular  AM  drawn  from  A  to  BC;  the 
ratio  of  BA  to  AC  must  be  less  than  (8.  5.)  the  ratio  of  BA  to 
AM,  because  AC  is  greater  than  AM.  Make  as  BA  to  AC,  so 
ED  to  DH;  therefore  the  ratio  of  ED  to  DH  is  less  than  the  ratio 
of  (BA  to  AM,  that  is,  than  the  ratio  of)  ED  to  DG;  and  conse- 
quently DH  is  greater  (10.  5.)  than  DG;  and  because  BA  is  great- 


EUCLID  S    DATA. 


323 


er  than  AC,  ED  is  greater  (A.  5.)  than 
DH.     From   the   centre  D,   at  the  dis- 
tance   DH,    describe    the    circle    KHF 
which  necessarily  meets  the  straight  line 
EF  in  two  points,  because  DH  is  great- 
er than  DG,  and  less  than  DE.     Let  the 
circle  meet  EF  in  the  points  F,  K  which 
are  given,  as  was  shown  in  the  preced- 
ing case,-  and  DF,  DK  being  joined,  the 
triangles  DEF,  DEK  are  given  in  spe- 
cies, as  was  there  shown.     From  the  cen- 
tre A,   at  the   distance   AC,   describe  a 
circle  meeting  BC  again  in  L:  and  if  the 
angle  ACB  be  less  than  a  right  angle,  ALB  must  be  greater  than 
a  right  angle;  and  on  the  contrary.     In  the  same  manner,  if  the 
angle  DGF  be  less  than  aright  angle,  DKE  must  be  greater  than 
one;    and  on  the  contrary.     Let   each  of 
the  angles  ACB,  DFE  be  either  less  or  A 

greater  than  a  right  angle;  and  because  in 
the  triangles  ABC,  DEF,  the  angles  ABC, 
DEF  are  equal,  and  the  sides  BA,  AC 
and  ED,  DF  about  two  of  the  other 
angles  proportionals,  the  triangle  ABC 
is  similar  (7.  6.)  to  the  triangle  DEF. 
In  the  same  manner,  the  triangle  ABL 
is  similar  to  DEK.  And  the  triangles 
DEF,  DEK  are  given  in  species;  there- 
fore also  the  trianeles  ABC,  ABL  are 
given  in  species.  And  from  this  it  is 
evident,  that  in  this  third  case  there  are 

always  two  triangles  of  a  different  species,  to  which  the  things 
mentioned  as  given  in  the  proposition  can  agree. 


PROP.  XLVIII. 


45. 


If  a  triangle  has  one  angle  given,  and  if  both  the  sides 
together  about  that  angle  have  a  given  ratio  to  the  re- 
maining side;  the  triangle  is  given  in  species. 

Let  the  triangle  ABC  have  the  angle  BAC  given,  and  let  the 
sides  BA,  AC  together  about  that  angle  have  a  given  ratio  to 
BC;  the  triangle  ABC  is  given  in  species. 

Bisect  (9.  1.)  the  angle  BAC  by  the  straight  line  AD;  there- 
fore the  angle  BAD  is  given.     And  because  as  BA  to  AC,  so  is 
(3.  6.)  BD  to  DC;  by  permutation,  as  AB  to 
BD,  so  is  AC  to  CD:  and  as  BA  and  AC  to-  A 

gether  to  BC,  so  is  (12.  5.)  AB  to  BD.  But 
the  ratio  of  BA  and  AC  together  to  BC 
is  given,  wherefore  the  ratio  of  AB  to  BD  is 
given,  and  the  angle  BAD  is  given:  therefore 
(47.  dat.)  the  triangle  ABD  is  given  in  spe- 
cies, and  the  angle  ABD  is  therefore  given:  the  angle  BAC  is  also 
given:  wherefore  the  triangle  ABC  is  given  in  species  (43.  dat.). 


324  Euclid's  data. 

A  triangle  which  shall  have  the  things  that  are  mentioned  in 
the  proposition  to  be  given,  can  be  found  in  the  following  man- 
ner. Let  EFG  be  the  given  angle,  and  let  the  ratio  of  H  to  K  be 
the  given  ratio  which  the  two  sides  about  the  angle  EFG  must 
have  to  the  third  side  of  the  triangle:  therefore,  because  two  sides 
of  a  triangle  are  greater  than  the  third  side,  the  ratio  of  H  to  K 
must  be  tlie  ratio  of  a  greater  to  a  less.  Bisect  (9.  1.)  the  angle 
EFG  by  the  straight  line  FL,  and  by  the  47th  proposition  find  a. 
triangle  of  which  EFL  is  one  of  the  angles,  and  in  which  the  ratio 
of  the  sides  about  the  angle  opposite  to  FL  is  the  same  with  the 
ratio  of  H  to  K:  to  do  which  take  FE  given  in  position  and  mag- 
nitude, and  draw  EL  perpendicular  to  FL;  then,  if  the  ratio  of  H 
to  K  be  the  same  with  the  ratio  of  FE  to  EL,  produce  EL,  and  let 
it  meet  EG  in  P:  the  triangle  FEP  is  that  which  was  to  be  found: 
For  it  has  the  given  angle  EFG;  and  because  this  angle  is  bi- 
sected by  FL,  the  sides  EF,  FP  together  are  to  EP,  as  (3.  6.)  FE 
to  EL,  that  is,  as  H  to  K. 

But  if  the  ratio  of  H  to  K  be  not  the  same  with  the  ratio  of  FE 
to  EL,  it  must  be  less  than  it,  as  was  shown  in  prop.  47,  and  in 
this  case  there  are  two  triangles,  each 
of  which  has  the  given  angle  EFL, 
and  the  ratio  of  the  sides  about  the 
angle  opposite  to  FL  the  same  with 
the  ratio  of  H  to  K.  By  prop.  47, 
find  these  triangles  EFM,  EFN,  each 
of  which  has  the  angle  EFL  for  one 
of  its  angles,  and  the  ratio  of  the  side 
F"E  to  EM  or  EN  the  same  with  the 
ratio  of  H  to  K;  and  let  the  angle  EMF  be  greater,  and  ENF  less 
than  a  right  angle.  And  because  H  is  greater  than  K,  EF  is 
greater  than  EN,  and  therefore  the  angle  EFN,  that  is,  the  angle 
NFG,  is  less  (18.  1.)  than  the  angle  ENF.  To  each  of  these  add 
the  angles  NEF,  EFN:  therefore  the  angles  NEF,  EFG  are  less 
than  the  angles  NEF,  EFN,  FNE,  that  is,  than  two  right  angles: 
therefore  the  straight  lines  EN,  EG  must  meet  together  when 
produced;  let  them  meet  in  O,  and  produce  EM  to  G.  Each  of 
the  triangles  EFG,  EFO  has  the  things  mentioned  to  be  given  in 
the  proposition:  for  each  of  them  has  the  given  angle  EFG;  and 
because  this  angle  is  bisected  by  the  straight  line  FMN,  the  sides 
EF,  EG  together  have  to  EG  the  third  side  the  ratio  of  FE  to 
EM,  that  is,  of  H  to  K.  In  like  manner,  the  sides  EF,  FO  toge- 
ther have  to  EO  the  ratio  which  H  has  to  K. 

PROP.  XLIX.  46. 

If  a  triangle  has  one  angle  given,  and  if  the  sides 
about  another  angle,  both  together,  have  a  given  ratio 
to  the  third  side;  the  triangle  is  given  in  species. 

Let  the  triangle  ABC  have  one  angle  ABC  given,  and  let  the 
two  sides  BA,  AC  about  another  angle  BAG  have  a  given  ratio 
to  BC;  the  triangle  ABC  is  given  in  species. 


Euclid's  data.  325 

Suppose  the  angle  BAG  to  be  bisected  by  the  straight  line  AD: 
BA  and  AC  together  are  to  BC,  as  AB  to  BD,  as  was  shown  in 
the  preceding  proposition.  But  the  ratio  of  BA  and  AC  together 
to  BC  is  given,  therefore  also  the  ratio  of  AB  to  BD  is  given. 
And  the  angle  ABD  is  given,  wherefore  (44.  dat.)  the  triangle 
ABD  is  given  in  species:  and  consequently  the  angle  BAD,  and 
its  double  the  angle  BAC  are  given; 
and  the  angle  ABC  is  given.  There- 
fore the  triangle  ABC  is  given  in  spe- 
cies (43.  dat.). 

A  triangle  which  shall  have  the  things 
mentioned  in  the  proposition  to  be  given, 
may  be  thus  found.  Let  EFG  be  the 
given  angle,  and  the  ratio  of  H  to  K  the 
given  ratio:  and  by  prop.  44,  find  the 
triangle  EFL,  which  has  the  angle  EFG 
for  one  of  its  angles,  and  the  ratio  of  the 
sides  EF,  FL  about  this  angle  the  same 
with  the  ratio  of  H  to  K;  and  make  the 
angle  LEM  equal  to  the  angle  FEL.  And  because  the  ratio  of 
H  to  K  is  the  ratio  which  two  sides  of  a  triangle  have  to  the  third, 
H  must  be  greater  than  K;  and  because  EF  is  to  FL,  as  H  to  K, 
therefore  EF  is  greater  than  FL,  and  the  angle  FEL,  that  is, 
LEM,  is  therefore  less  than  the  angle  ELF.  Wherefore  the  an- 
gles LFE,  FEM  are  less  than  two  right  angles,  as  was  shown  in 
the  foregoing  proposition,  and  the  straight  lines  FL,  EM  must 
meet,  if  produced:  let  them  meet  in  G, EFG  is  the  triangle  which 
was  to  be  found;  for  EFG  is  one  of  its  angles,  and  because  the 
angle  EFG  is  bisected  by  EL,  the  two  sides  FE,  EG  together 
have  to  the  third  side  EG  the  ratio  of  EF  to  FL,  that  is,  the  given 
ratio  of  H  to  K. 

PROP.  L.  76. 

If  from  the  vertex  of  a  triangle,  given  in  species,  a 
straight  Hne  be  drawn  to  the  base  in  a  given  angle,  it 
shall  have  a  given  ratio  to  the  base. 

From  the  vertex  A  of  the  triangle  ABC  which  is  given  in  spe- 
cies, let  AD  be  drawn  to  the  base  BC  in  a  given  angle  ADB;  the 
ratio  of  AD  to  BC  is  given. 

Because  the  triangle  ABC  is  given  in  spe- 
cies, the  angle  ABD  is  given,  and  the  angle 
ADB  is  given,  therefore  the  triangle  ABD  is 
given  (43.  dat.)  in  species;  wherefore  the  ratio 
of  AD  to  AB  is  given.  And  the  ratio  of  AB 
to  BC  is  given;  and  therefore  (9.  dat.)  the 
ratio  of  AD  to  BC  is  given. 

PROP.  LL  47. 

Rectilineal  figures,  given  in  species,  are  divided 
into  triangles  which  are  given  in  species. 


326 


EUCLID  S  DATA. 


B 


Let  the  rectilineal  figure  ABCDE  be  given  in  species;  ABCDE 
may  be  divided  into  triangles  given  in  species. 

Join  BE,  BD;  and  because  ABCDE  is  given  in  species,  the 
angle  BAE  is  given  (3.  def.),  and  the  ratio 
of  BA  to  AE  is  given  (3.  def,);  wherefore 
the  triangle  BAE  is  given  in  species  (44. 
dat.),  and  the  angle  AEB  is  therefore  given 
(3.  def.).  But  the  v^^hole  angle  AED  is 
given,  and  therefore  the  remaining  angle 
BED  is  given,  and  the  ratio  of  AE  to  EB 
is  given,  as  also  the  ratio  of  AE  to  ED;  C  D 

therefore  the  ratio  of  BE  to  ED  is  given  (9.  dat.).  And  the  angle 
BED  is  given,  wherefore  the  triangle  BED  is  given  (44.  dat.)  in 
species.  In  the  same  manner,  the  triangle  BDC  is  given  in  spe- 
cies: therefore  rectilineal  figures  which  are  given  in  species  are 
divided  into  triangles  given  in  species. 


PROP.  LIL 


48. 


If  two  triangles  given  in  species  be  described  upon 
the  same  straight  hne,  they  shall  have  a  given  ratio  to 
one  another. 

Let  the  triangles  ABC,  ABD,  given  in  species,  be  described 
upon  the  same  straight  line  AB;  the  ratio  of  the  triangle  ABC  to 
the  triangle  ABD  is  given. 

Through  the  point  C  draw  CE  parallel  to  AB,  and  let  it  meet 
DA  produced  in  E,  and  join  BE.  Because  the  triangle  ABC  is 
given  in  species,  the  angle  BAC,  that  is,  the  angle  ACE,  is  given; 
and  because  the  triangle  ABD  is  given  in  species,  the  angle  DAB, 
that  is,  the  angle  AEC       E  C 

is  given.  Therefore  the 
triangle  ACE  is  given 
in  species;  wherefore 
the  ratio  of  EA  to  AC 
is  given  (3.  def.),  and 
the  ratio  of  CA  to  AB 
is  given,  as  also  the  ra- 
tio of  BA  to  AD;  there- 
fore the  ratio  of  (9.  dat.)  EA  to  AD  is  given,  and  the  triangle 
ACB  is  equal  (37.  1.)  to  the  triangle  AEB,  and  as  the  triangle 
AEB,  or  ACB,  is  to  the  triangle  ADB,  so  is  (1.  6.)  the  straight 
line  EA  to  AD.  But  the  ratio  of  EA  to  AD  is  given,  therefore 
the  ratio  of  the  triangle  ACB  to  the  triangle  ADB  is  given. 

PROBLEM. 

To  find  the  ratio  of  two  triangles  ABC,  ABD  given  in  species, 
and  which  are  described  upon  the  same  straight  line  AB. 

Take  a  straight  line  FG  given  in  position  and  magnitude,  and 
because  the  angles  of  the  triangles  ABC,  ABD  are  given,  at  the 
points  F,  G  of  the  straight  line  FG,  make  the  angles  GFH,  GFK 
(23.  1.)  equal  to  the  angles  BAC,  BAD:  and  the  angles  FGH, 


EUCLID  S  DATA. 


327 


FGK  equal  to  the  angles  ABC,  ABD,  each  to  each.  Therefore 
the  triangles  ABC,  ABD  are  equiangular  to  the  triangles  FGH, 
FGK,  each  to  each.  Through  the  point  H  draw  HL  parallel  to 
FG,  meeting  KF  produced  in  L.  And  because  the  angles  BAG, 
BAD  are  equal  to  the  angles  GFH,GFK,  each  to  each;  therefore 
the  angles  ACE,  AEC  are  equal  to  FHL,  FLH,  each  to  each,  and 
the  triangle  AEC  equiangular  to  the  triangle  FLH.  Therefore 
as  EA  to  AC,  so  is  LF  to  FH;  and  as  CA  to  AB,  so  EIF  to  FG5 
and  as  BA  to  AD,  so  is  GF  to  FK;  wherefore,  ex  sequali^  as  EA 
to  AD,  so  is  LF  to  FK.  But,  as  was  shown,  the  triangle  ABC  is 
to  the  triangle  ABD,  as  the  straight  line  EA  to  AD,  that  is,  as 
LF  to  FK.  The  ratio  therefore  of  LF  to  FK  has  been  found, 
which  is  the  same  with  the  ratio  of  the  triangle  ABC  to  the  tri- 
angle ABD. 


PROP.  Lin. 


49. 


If  two  rectilineal  figures  given  in  species  be  described 
upon  the  same  straight  line,  they  shall  have  a  given 
ratio  to  one  another.^ 

Let  any  two  rectilineal  figures  ABCDE,  ABFG,  which  are 
given  in  species,  be  described  upon  the  same  straight  line  ABj 
the  ratio  of  them  to  one  another  is  given. 

Join  AC,  AD,  AF:  each  of  the  triangles  AED,  ADC,  ACB, 
AGF,  ABF  is  given  (51.  dat.)  in  species.  And  because  the  trian- 
gles ADE,  ADC  given  in  species  are 
described  upon  the  same  straight  line 
AD,  the  ratio  of  EAD  to  DAC  is 
given  (52.  dat.);  and,  by  composition, 
the  ratio  of  EACD  to  DAC  is  given 
(7.  dat.).  And  the  ratio  of  DAC  to 
CAB  is  given  (52.  dat.)  because  they 
are  described  upon  the  same  straight 
line  AC;  therefore  the  ratio  of  Ex\CD 
to  ACB  is  given  (9.  dat.);  and,  by 
composition,  the  ratio  of  ABCDE  to 
to  ABC  is  given.  In  the  same  man- 
ner, the  ratio  of  ABFG  to  ABF  is  given.  But  the  ratio  of  the 
triangle  ABC  to  the  triangle  ABF  is  given;  wherefore  (52.  dat), 
because  the  ratio  of  ABCDE  to  ABC  is  given,  as  also  the  ratio 
of  ABC  to  ABF,  and  the  ratio  of  ABF  to  ABFG;  the  ratio  of  the 
rectilineal  ABCDE  to  the  rectilineal  ABFG  is  given  (9.  dat.). 

PROBLEM. 

To  find  the  ratio  of  two  rectilineal  figures  given  in  species,  and 
described  upon  the  same  straight  line. 

Let  ABCDE,  ABFG  be  two  rectilineal  figures  given  in  species, 
and  described  upon  the  same  straight  line  AB,  and  join  AC,  AD, 
AF.  Take  a  straight  line  HK  given  in  position  and  magnitude, 
and  by  the  52d  dat.  find  the  ratio  of  the  triangle  ADE  to  the  tri- 


H- 


K  LM  N 

—I— I— 1—0 


See  Note. 


328  Euclid's  data. 

angle  ADC,  and  make  the  ratio  of  HK  to  KL  the  same  with  it. 
Find  also  the  ratio  of  the  triangle  ABD  to  the  triangle  ACB. 
And  make  the  ratio  of  KL  to  LM  the  same.  Also,  find  the  ratio 
of  the  triangle  ABC  to  the  triangle  ABF,  and  make  the  ratio  of 
LM  to  MN  the  same.  And,  lastly,  find  the  ratio  of  the  triangle 
AFB  to  the  triangle  AFG,  and  make  ~ 

the  ratio  of  MN  to  NO  the  same. 
Then  the  ratio  of  ABCDE  to  ABFG 
is  the  same  with  the  ratio  of  HM  to 
MO. 

Because  the  triangle  EAD  is  to 
the  triangle  DAC  as  the  straight 
line  HK  to  KL;  and  as  the  triangle 
DAC  to  CAB,  so  is  the  straight  line 
KL  to  LM;  therefore,  by  using  com-  K  L  M  N 

position  as  often  as  the  number  of  H ] — -| — [ — | — O 

triangles    requires,    the    rectilineal 

ABCDE  is  to  the  triangle  ABC,  as  the  straight  line  HM  to  ML. 
In  like  manner,  because  the  triangle  GAF  is  to  FAB,  as  ON  to 
NM,  by  composition,  the  rectilineal  ABFG  is  to  the  triangle 
ABF,  as  MO  to  NM;  and,  by  inversion,  as  ABF  to  ABFG,  so  is 
NM  to  MO.  And  the  triangle  ABC  is  to  ABF,  as  LM  to  MN. 
Wherefore,  because  as  ABCDE  to  ABC,  so  is  HM  to  ML;  and 
as  ABC  to  ABF,  so  is  LM  to  MN;  and  as  ABF  to  ABFG,  so  is 
MN  to  MO:  ex  sequali,  as  the  rectilineal  ABCDE  to  ABFG,  so 
is  the  straight  line  HM  to  MO. 

PROP.  LIV.  50. 

If  two  straight  lines  have  a  given  ratio  to  one  ano- 
ther, the  similar  rectilineal  figures  described  upon  them 
similarly,  shall  have  a  given  ratio  to  one  another. 

Let  the  straight  lines  AB,  CD  have  a  given  ratio  to  one  ano- 
ther, and  let  the  similar  and  similarly  placed  rectilineal  figures 
E,  F  be  described  upon  them;  the  ratio  of  E  to  F  is  given. 

To  AB,  CD,  let  G  be  a  third  propor- 
tional: therefore,  as  AB  to  CD,  so  is  CD 
to  G.  And  the  ratio  of  AB  to  CD  is 
given,  wherefore  the  ratio  of  CD  to  G  is 
given;  and  consequently  the  ratio  of  AB 
to  G  is  also  given  (9.  dat.).  But  as  AB 
to  G,  so  is  the  figure  E  to  the  figure  (2.  H  K        L 

cor.  20.  6.)  F.     Therefore  the  ratio  of  E        — 

to  F  is  given. 

PROBLEM. 

To  find  the  ratio  of  two  similar  rectilineal  figures,  E,  F,  simi- 
larly described  upon  straight  lines  AB,  CD  which  have  a  given 
ratio  to  one  another:  let  G  be  a  third  proportional  to  AB,  CD. 

Take  a  straight  line  H  given  in  magnitude;  and  because  the 
ratio  of  AB  to  CD  is  given,  make  the  ratio  of  H  to  K  the  same 


F 


Euclid's  data.  329 

with  it;  and  because  H  is  given,  K  is  given.  As  H  is  to  K,  so 
make  K  to  L;  then  the  ratio  of  E  to  F  is  the  same  with  the  ratio 
of  H  to  L:  for  AB  is  to  CD,  as  H  to  K,  wherefore  CD  is  to  G,  as 
K  to  L:  and,  ex  eequali^  as  A B  to  G,  so  is  H  to  L:  but  the  figure 
E  is  to  (2.  cor.  20.  6.)  the  figure  F,  as  AB  to  G,  that  is,  as  H  to  L. 

PROP.  LV.  51. 

If  two  straight  lines  have  a  given  ratio  to  one  ano- 
ther; the  rectihneal  figures  given  in  species  described 
upon  them,  shall  have  to  one  another  a  given  ratio. 

Let  AB,  CD  be  two  straight  lines  which  have  a  given  ratio  to 
one  another:  the  rectilineal  figures  E,  F  given  in  species  and  de- 
scribed upon  them,  have  a  given  ratio  to  one  another. 

Upon  the  straight  line  AB,  describe  the  figure  AG  similar  and 
similarly  placed  to  the  figure  F;  and  because  F  is  given  in  spe- 
cies, AG  is  also  given  in  species: 
therefore,  since  the  figures  E,  AG, 

which  are  given  in  species,  are  X t?^\  ^  ^ 
described  upon  the  same  straight 
line  AB,  the  ratio  of  E  to  AG  is 
given  (53.  dat.),  and  because  the 
ratio  of  AB  to  CD  is  given,  and 
upon  them  are  described  the  si- 
milar and  similarly  placed  recti-         H K L 

lineal  figures  AG,  F,  the  ratio  of 

AG  to  F  is  given  (54.  dat.):  and  the  ratio  of  AG  to  E  is  given: 

therefore  the  ratio  of  E  to  F  is  given  (9.  dat.). 

PROBLEM. 

To  find  the  ratio  of  two  rectilineal  figures  E,  F  given  in  species, 
and  described  upon  the  straight  lines  AB,  CD  which  have  a  given 
ratio  to  one  another. 

Take  a  straight  line  H  given  in  magnitude;  and  because  the 
rectilineal  figures  E,  AG  given  in  species  are  described  upon  the 
same  straight  line  AB,  find  their  ratio  by  the  53d  dat.  and  make 
the  ratio  of  H  to  K  the  same;  K  is  therefore  given:  and  because 
the  similar  rectilineal  figures  AG,  F  are  described  upon  the 
straight  lines  AB,  CD,  which  have  a  given  ratio,  find  their  ratio 
by  the  54th  dat.  and  make  the  ratio  of  K  to  L  the  same:  the  figure 
E  has  to  F  the  same  ratio  which  H  has  to  L:  for  by  the  construc- 
tion, as  E  is  to  AG,  so  is  H  to  K;  and  as  AG  to  F,  so  is  K  to  L; 
therefore,  ex  sequali,  as  E  to  F,  so  is  H  to  L. 

PROP.  LVL  52. 

If  a  rectilineal  figure  given  in  species  be  described 
upon  a  straight  line  given  in  magnitude,  the  figure  is 
given  in  magnitude. 

Let  the  rectilineal  figure  ABCDE  given  in  species  be  described 
2  T 


330 


EUCLID  S  DATA. 


Upon  the  straight  line  AB  given  in  magnitudej  the  figure  ABCDE 
is  given  in  magnitude.  ~    . 

Upon  AB  let  the   square  AF  be   de-  C 

scribed^  therefore  AF  is  given  in  species 
and  magnitude,  and  because  the  rectilineal 
figures  ABCDE,  AF  given  in  species  are  j^ 
described  upon  the  same  straight  line  AB, 
the  ratio  of  ABCDE  to  AF  is  given  (53. 
dat.):  but  the  square  AF  is  given  in  mag- 
nitude, therefore  (2.  dat.)  also  the  figure 
ABCDE  is  given  in  magnitude. 

PROBLEM. 

To  find  the  magnitude  of  a  rectilineal 
figure  given  in  species  described  upon  a 
straight  line  given  in  magnitude. 

Take  the  straight  line  GH  equal  to  the 
given  straight  line  AB,  and  by  the  53d  dat. 
find  the  ratio  which  the  square  AF  upon 
AB  has  to  the  figure  ABCDEj  and  make  the  ratio  of  GH  to  HK 
the  samej  and  upon  GH  describe  the  square  GL,  and  complete 
the  parallelogram  LHKM;  the  figure  ABCDE  is  equal  to  LHKM: 
because  AF  is  to  ABCDE,  as  the  straight  line  GH  to  HK,  that 
is,  as  the  figure  GL  to  HM;  and  AF  is  equal  to  GL;  therefore 
ABCDE  is  equal  to  HM  (14.  5.). 

PROP.  LVH.  53. 

If  two  rectilineal  figures  are  given  in  species,  and  if 
a  side  of  one  of  them  has  a  given  ratio  to  a  side  of  the 
other ;  the  ratios  of  the  remaining  sides  to  the  remain- 
ing sides  shall  be  given. 

Let  AC,  DF,  be  two  rectilineal  figures  given  in  species,  and  let 
the  ratio  of  the  side  AB  to  the  side  DE  be  given,  the  ratios  of  the 
remaining  sides  to  the  remaining  sides  are  also  given. 

Because  the  ratio  of  AB  to  DE  is  given,  as  also  (3.  def.)  the 
ratios  of  AB  to  BC,  and  of  DE  to  EF,  the  ratio  of  BC  to  EF  is 
given  (10.  dat.).  In  the  same  manner,  the 
ratios  of  the  other  sides  to  the  other  sides  . 
are  given. 

The  ratio  which  BC  has  to  EF  may  be 
found  thus:  take  a  straight  line  G  given  in 
magnitude,  and  because  the  ratio  of  BC  to 
BA  is  given,  make  the  ratio  of  G  to  H  the 
same;  and  because  the  ratio  of  AB  to  DE 
is  given,  make  the  ratio  of  H  to  K  the  same; 
and  make  the  ratio  of  K  to  L  the  same  with 
the  given  ratio  of  DE  to  EF.  Since  there- 
fore as  BC  to  BA,  so  is  G  to  H;  and  as  BA 
to  DE,  so  is  H  to  K;  and  as  DE  to  EF,  so  is  K  to  L;  ex  iequali^  BC 
is  to  EF,  as  G  to  L;  therefore  the  ratio  of  G  to  L  has  been  found, 
which  is  the  same  with  the  ratio  of  BC  to  EF. 


H      K      L 


EUCLID  S  DATA. 


331 


H 


E      F  G 


K 


PROP.  LVIII.  G. 

If  two  similar  rectilineal  figures  have  a  given  ratio 
to  one  another,  their  homologous  sides  have  also  a 
given  ratio  to  one  another.* 

Let  the  two  similar  rectilineal  figures,  A,  B,  have  a  given  ratio 
to  one  another,  their  homologous  sides  have  also  a  given  ratio. 

Let  the  side  CD  be  homologous  to  EF,  and  to  CD,  EF  let 
the  straight  line  G  be  a  third  proportional.  As  therefore  (2.  cor. 
20.  C.)  CD  to  G,  so  is  the  figure  A  to  B; 
and  the  ratio  of  A  to  B  is  given,  therefore 
the  ratio  of  CD  to  G  is  givenj  and  CD, 
EF,  G  are  proportionals;  wherefore  (13. 
dat.)  the  ratio  of  CD  to  EF  is  given. 

The  ratio  of  CD  to  EF  may  be  found 
thus:  Take  a  straight  line  H  given  in 
magnitude^  and  because  the  ratio  of  the 
figure  A  to  B  is  given,  make  the  ratio  of  H  to  K  the  same  with  it: 
And,  as  the  13th  dat,  directs  to  be  done,  find  a  mean  proportional 
L  between  H  and  K;  the  ratio  of  CD  to  EF  is  the  same  with  that 
of  H  to  L.  Let  G  be  a  third  proportional  to  CD,  EF;  therefore  as 
CD  to  G,  so  is  (A  to  B,  and  so  is)  H  to  K;  and  as  CD  to  EF,  so 
is  H  to  L,  as  is  shown  in  the  13th  dat. 

PROP.  LIX.  54. 

If  two  rectilineal  figures  given  in  species  have  a 
given  ratio  to  one  another,  their  sides  shall  likewise 
have  given  ratios  to  one  another.* 

Let  the  two  rectilineal  figures  A,  B,  given  in  species,  have  a 
given  ratio  to  one  another,  their  sides  shall  also  have  given  ratios 
to  one  another. 

If  the  figure  A  be  similar  to  B,  their  homologous  sides  shall 
have  a  given  ratio  to  one  another,  by  the  preceding  proposition; 
and  because  the  figures  are  given  in  species,  the  sides  of  each  of 
them  have  given  ratios  (3.def.)  to  one  another;  therefore  each  side 
of  two  of  them  has  (9.  dat.)  to  each  side  of  the  other  a  given  ratio. 

But  if  the  figure  A  be  not  similar  to  B,  let  CD,  EF  be  any  two 
of  their  sides;  and  upon  EF  conceive  the  figure  EG  to  be  described 
similar  and  similarly  placed  to  the 
figure  A,  so  that  CD,  EF  be  homo- 
logous sides;  therefore  EG  is  given 
in  species;  and  the  figure  B  is  given 
in  species;  wherefore  (53.  dat.)  the 
ratio  of  B  to  EG  is  given;  and  the 
ratio  of  A  to  B  is  given,  therefore 
(9.  dat.)  the  ratio  of  the  figure  A  to 
EG  is  given;  and  A  is  similar  to 
EG;  therefore  (58.  dat.)  the  ratio 
of  the  side  CD  to  EF  is  given;  and 
consequently  (9.  dat.)  the  ratios  of 
the  remaining  sides  to  the  remaining  sides  are  given. 

^  See  Note. 


H- 
K- 
M- 
L- 


332 


EUCLID  S  DATA. 


-  L 


The  ratio  of  CD  to  EF  may  be  found  thus:  take  a  straight  line 
H  given  in  magnitude,  and  because  the  ratio  of  the  figure  A  to  B 
is  given,  make  the  ratio  of  H  to  K  the  same  with  it.  And  by  the 
53d  dat.  find  the  ratio  of  the  figure  B  to  EG,  and  make  the  ratio 
of  K  to  L  the  same:  Between  H  and  L  find  a  mean  proportional 
M;  the  ratio  of  CD  to  EF  is  the  same  with  the  ratio  of  H  to  M; 
because  the  figure  A  is  to  B  as  H  to  K;  and  as  B  to  EG,  so  is  K 
to  L;  ex  sequali,  as  A  to  EG  so  is  H  to  L:  and  the  figures  A,  EG  are 
similar,  and  M  is  a  mean  proportional  between  H  and  h;  there- 
fore, as  was  shown  in  the  preceding  proposition,  CD  is  to  EF  as 
H  to  M. 

PROP.  LX.  55. 

If  a  rectilineal  figure  be  given  in  species  and  mag- 
nitude, the  sides  of  it  shall  be  given  in  magnitude. 

Let  the  rectilineal  figure  A  be  given  in  species  and  magnitude, 
its  sides  are  given  in  magnitude. 

Take  a  straight  line  BC  given  in  position  and  magnitude,  and 
upon  BC  describe  (18.  6.)  the  figure  D  similar,  and  similarly 
placed,  to  the  figure  A,  and 
let  EF  be  the  side  of  the  figure 
A  homologous  to  BC  the  side 
of  D;  therefore  the  figure  D 
is  given  in  species.  And  be- 
cause upon  the  given  straight 
line  BC,  the  figure  D  given 
in  species  is  described,  D  is 
given  (56.  dat.)  in  magnitude, 
and  the  figure  A  is  given  in 
magnitude,  therefore  the  ra- 
tio of  A  to  D  is  given:  and  the  figure  A  is  similar  to  D^  there- 
fore the  ratio  of  the  side  EF  to  the  homologous  side  BC  is  given 
(58.  dat.);  wherefore  (2.  dat.)  EF  is  given:  and  the  ratio  of  EF 
to  EG  is  given  (3.  def.),  therefore  EG  is  given.  And,  in  the  same 
manner,  each  of  the  other  sides  of  a  figure  A  can  be  shown  to  be 
given. 

PROBLEM. 

To  describe  a  rectilineal  figure  A,  similar  to  a  given  figure  D, 
and  equal  to  another  given  figure  H.     It  is  prop.  25,  b.  6,  Elem. 

Because  each  of  the  figures  D,  H  is  given,  their  ratio  is  given, 
which  may  be  found  by  making  (cor.  45.  1.)  upon  the  given 
straight  line  BC  the  parallelogram  BK  equal  to  D,  and  upon  its 
side  CK  making  (cor.  45.  1.)  the  parallelogram  KL  equal  to  H, 
and  the  angle  KCL  equal  to  the  angle  MBC;  therefore  the  ratio 
of  D  to  H,  that  is,  of  BK  to  KL,  is  the  same  with  the  ratio  of  BC 
to  CL:  and  because  the  figures  D,  A  are  similar,  and  that  the 
ratio  of  D  to  A,  or  H,  is  the  same  with  the  ratio  of  BC  to  CL; 
by  the  58th  dat.  the  ratio  of  the  homologous  sides  BC,  EF  is  the 
same  with  the  ratio  of  BC  to  the  mean  proportional  between  BC 
and  CL.  Find  EF  the  mean  proportional;  then  EF  is  the  side  of 
the  figure  to  be  described,  homologous  to  BC  the  side  of  D,  and 


Euclid's  data.  333 

the  figure  ilself  can  be  described  by  the  18th  prop,  book  6,  which, 
by  the  construction,  is  similar  to  D;  and  because  D  is  to  A,  as 
(2  cor.  20.  6.)  BC  to  CL,  that  is,  as  the  figure  BK  to  KL;  and  that 
D  is  equal  to  BK,  therefore  A  (14.  5.)  is  equal  to  KL,  that  is,  to  H. 

PROP.  LXI.  57. 

If  a  parallelogram  given  in  magnitude  has  one  of  its 
sides  and  one  of  its  angles  given  in  magnitude,  the 
other  side  also  is  given.* 

Let  the  parallelogram  ABDC  given  in  magnitude,  have  the 
side  AB  and  the  angle  BAG  given  in  magnitude,  the  other  side 
AC  is  given. 

Take  a  straight  line  EF  given  in  position  and  magnitude;  and 

because  the  parallelogram  AD  is  given  in         A B 

magnitude,  a  rectilineal  figure  equal  to  it         /  "        7 

can  be  found  (l.  def.).     And  a  parallelo-        /  / 

gram  equal  to  this  figure  can  be  applied      / / 

(cor.  45.  1.)  to  the  given  straight  line  EF      c  D 

in  an  angle  equal  to  the  given  angle  BAG. 
Let  this  be  the  parallelogram  EFHG  hav- 
ing the  angle  FEG  equal  to  the  angle 
BAG.  And  because  the  parallelograms 
AD,  EH  are  equal  and  have  the  angles  at 
A  and  E  equal;  the  sides  about  them  are 
reciprocally  proportioned  (14.  6.);  there-     G  H 

fore  as  AB  to  EF,  so  is  EG  to  AC;  and  AB,  EF,  EG  are  given, 
therefore  also  AC  is  given  (12.  6.).  Whence  the  way  of  finding 
AC  is  manifest. 

PROP.  LXIL  H. 

If  a  parallelogram  has  a  given  angle,  the  rectangle 
contained  by  the  sides  about  that  angle  has  a  given 
ratio  to  the  parallelogram.* 

Let  the  parallelogram  ABCD  have  the  given 
angle  ABC,  the  rectangle  AB,  BC  has  a  given 
ratio  to  the  parallelogram  AC. 

From  the  point  A  draw  AE  perpendicular 

to  BC;  because  the  angle  ABC  is  given,  as  also  B  E 
the  angle  AEB,  the  triangle  ABE  is  given  (42. 
dat.)  in  species:  therefoi-e  the  ratio  of  BA  to 
AE  is  given.  But  as  BA  to  AE,  so  is  (l.  16.) 
the  rectangle  AB,  BC  to  the  rectangle  AE, 
BC;  therefore  the  ratio  of  the  rectangle  AB, 

BC  to  AE,  BC,  that  is,  (35.  1.)  to  the  parallel-    ^-^ H 

ogram  AC,  is  given. 

And  it  is  evident  how  the  ratio  of  the  rectangle  to  the  paral- 

*  See  Note. 


7 


334  Euclid's  data. 

lelog'ram  may  be  found  by  making  the  angle  FGH  equal  to  the 
given  angle  ABC,  and  drawing  from  any  point  F  in  one  of  its 
sides,  FK  perpendicular  to  the  other  GH:  for  OF  is  to  FK,  as 
BA  to  AE,  that  is,  as  the  rectangle  AB,  BC  to  the  parallelogram 
AC. 

Cor.  And  if  a  triangle  ABC  has  a  given  angle  ABC,  the  rect- 
angle AB,  BC  contained  by  the  sides  about  that  angle,  shall  have 
a  given  ratio  to  the  triangle  ABC. 

Complete  the  parallelogram  ABCD;  therefore,  by  this  propo- 
sition, the  rectangle  AB,  BC  has  a  given  ratio  to  the  parallelo- 
gram AC;  and  AC  has  a  given  ratio  to  its  half  the  triangle  (41. 
1.)  ABC;  therefore  the  rectangle  AB,  BC  has  a  given  (9.  dat.) 
ratio  to  the  triangle  ABC. 

And  the  ratio  of  the  rectangle  to  the  triangle  is  found  thus: 
make  the  triangle  FGK,  as  was  shown  in  the  proposition;  the 
ratio  of  GF  to  the  half  of  the  perpendicular  FK  is  the  same  with 
the  ratio  of  the  rectangle  AB,  BC  to  the  triangle  ABC.  Because, 
as  was  shown,  GF  is  to  FK,  as  AB,  BC  to  the  parallelogram  AC; 
and  FK  is  to  its  half,  as  AC  is  to  its  half,  which  is  the  triangle 
ABC;  therefore,  ex  sequali^  GF  is  to  the  half  of  FK,  as  AB,  BC 
rectangle  is  to  the  triangle  ABC. 

PROP.  LXIII.  56. 

If  two  parallelograms  be  equiangular,  as  a  side  of 
the  first  to  a  side  of  the  second,  so  is  the  other  side  of 
the  second  to  the  straight  line  to  which  the  other  side 
of  the  first  has  the  same  ratio  which  the  first  parallelo- 
gram has  to  the  second.  And  consequently,  if  the 
ratio  of  the  first  parallelogram  to  the  second  be  given, 
the  ratio  of  the  other  side  of  the  first  to  that  straight 
line  is  given,  and  if  the  ratio  of  the  other  side  of  the 
first  to  that  straight  line  be  given,  the  ratio  of  the  first 
parallelogram  to  the  second  is  given. 

Let  AC,  DF  be  two  equiangular  parallelograms;  as  BC,  a  side 
-of  the  first,  is  to  EF,  a  side  of  the  second,  so  is  DE,  the  other  side 
of  the  second,  to  the  straight  line  to  which  AB,  the  other  side  of 
the  first,  has  the  same  ratio  which  AC  has  to  DF. 

Produce  the  straight  line  AB,  and  make  as  BC  to  EF,  so  DEto 
BG,    and   complete    the    parallelogram  A 

BGHC;  therefore  because  BC  or  GH  is 
to  EF,  as  DE  to  BG,  the  sides  about  the 
equal  angles  BGH,DEF  are  reciprocally 
proportional;  wherefore  (14.  6.)  the  pa- 
rallelogram BH  is  equal  to  DF;  and  AB 
is  to  BG,  as  the  parallelogram  AC  is  to 
BH,  that  is,  to  DF;  as  therefore  BC  is 
to  EF,  so  is  DE  to  BG,  which  is  the 
straight  line  to  which  AB  has  the  same 
ratio  that  AC  has  to  DF. 


-r/ 

/ 

/c 

-"/ 

r 

G 

D 

H 

F,/ 

/ 

F 

EUCLID*S  DATA.  335 

And  if  the  ratio  of  the  parallelogram  AC  to  DF  be  |iven,  then 
the  ratio  of  the  straight  line  AB  to  BG  is  given;  and  if  the  ratio  of 
AB  to  the  straight  line  BG  be  given,  the  ratio  of  the  parallelo- 
gram AC  to  DF  is  given. 

PROP.  LXIV.  74.  73. 

iptwo  parallelograms  have  unequal  but  given  an- 
gles, and  if  as  a  side  of  the  first  to  a  side  of  the  second, 
so  the  other  side  of  the  second  be  made  to  a  certain 
straight  line;  if  the  ratio  of  the  first  parallelogram  to 
the  second  be  given,  the  ratio  of  the  other  side  of  the 
first  to  that  straight  line  shall  be  given.  And  if  the 
ratio  of  the  other  side  of  the  first  to  that  straight  line 
be  given,  the  ratio  of  the  first  parallelogram  to  the  se- 
cond shall  be  given.* 

Let  ABCD,  EFGH  be  two  parallelograms  vv^hich  have  the  un- 
equal, but  given  angles  ABC,  EFG;  and  as  BC  to  FG,  so  make 
EF  to  the  straight  line  M.  If  the  ratio  of  the  parallelogram  AC 
to  EG  be  given,  the  ratio  of  AB  to  M  is  given. 

At  the  point  B  of  the  straight  line  BC  make  the  angle  CBK 
equal  to  the  angle  EFG,  and  complete  the  parallelogram  KBCL. 
And  because  the  ratio  of  AC  to  EG  is  given,  and  that  AC  is  equal 
(35.  1.)  to  the  parallelogram  KC,  therefore  the  ratio  of  KC  to  EG 
is  given;  and  KC,  EG  are  equiangular;  therefore  as  BC  to  FG,  so 
is  (63.  dat.)  EF  to  the  straight  line  to  which  KB  has  a  given  ratio, 
viz.  the  same  which  the  parallelogram  KC  has  to  EG;  but  as  BC 
to  FG,  so  is  EF  to  the  straight  line  M;  therefore  KB  has  a  given 
ratio  to  M;  and  the  ratio  of  AB  to  BK  is  given,  because  the  trian- 
gle ABK  is  given  in  species  (43.  dat.);  therefore  the  ratio  of  AB 
to  M  is  given  (9.  dat.). 

And  if  the  ratio  of  AB  to  M  be  given,  the  ratio  of  the  parallelo- 
gram AC  to  EG  is  given;  for  since  the  ratio  of  KB  to  BA  is 
given,  as  also  the  ratio  of  AB  to  M,  the  K    A  L     D 

ratio  of  KB  to  M  is  given  (9.  dat.);  and 
because  the  parallelograms  KC,  EG  are 
equiangular,  as  BC  to  FG,  so  is  (63.  dat.) 
EF  to  the  straight  line  to  which  KB  has 
the  same  ratio  which  the  parallelogram 

KC  has  to  EG;  but  as  BC  to  FG,  so  is  

EF  to  M;  therefore  KB  is  to  M,  as  the  jyt    -g^ 

parallelogram  KC  is  to  EG;  and  the  ratio 

of  KB  to  M  is  given,  therefore  the  ratio  of 

the  parallelogram  KC,  that  is,  of  AC  to  EG,  is  given. 

Cor.  And  if  two  triangles  ABC,  EFG,  have  two  equal  angles, 
or  two  unequal,  but  given  angles,  ABC,  EFG,  and  if  as  BC  a  side 
of  the  first  to  FG  a  side  of  the  second,  so  the  other  side  of  the  se- 

*  See  Note. 


336  Euclid's  data. 

cond  EF  be  made  to  a  straight  line  M;  if  the  ratio  of  the  triangles 
be  given,  the  ratio  of  the  other  side  of  the  first  to  the  straight  line 
M  is  given. 

Complete  the  parallelograms  ABCD,  EFGH;  and  because  the 
ratio  of  the  triangle  ABC  to  the  triangle  EFG  is  given,  the  ratio 
of  the  parallelogram  AC  to  EG  is  given  (15.  5.),  because  the  pa- 
rallelograms are  double  (41.  1.)  of  the  triangles^  and  because  BC 
is  to  FG,  as  EF  to  M,  the  ratio  of  AB  to  M  is  given  by  the  63d 
dat.  if  the  angles  ABC,  EFG  are  equal;  but  if  they  be  unequal, 
but  given  angles,  the  ratio  of  AB  to  M  is  given  by  this  pro- 
position. 

And  if  the  ratio  of  AB  to  M  be  given,  the  ratio  of  the  paral- 
lelogram AC  to  EG  is  given  by  the  same  proposition;  and  there- 
fore the  ratio  of  the  triangle  ABC  to  EFG  is  given. 

PROP.  LXV.  68. 

If  two  equiangular  parallelograms  have  a  given  ratio 
to  one  another,  and  if  one  side  have  to  one  side  a  given 
ratio;  the  other  side  shall  also  have  to  the  other  side  a 
given  ratio. 

Let  the  tw^o  equiangular  parallelograms  AB,  CD  have  a  given 
ratio  to  one  another,  and  let  the  side  EB  have  a  given  ratio  to  the 
side  FD;  the  other  side  AE  has  also  a  given  ratio  to  the  other 
side  CF. 

Because  the  two  equiangular  parallelograms  AB,  CD  have  a 
given  ratio  to  one  another;  as  EB,  a  side  of  the  first,  is  to  FD,  a 
side  of  the  second,  so  is  (63.  dat.)  FC,  the  other  side  of  the  second, 
to  the  straight  line  to  which  AE,  the  other  side  of  the  first,  has 
the  same  given  ratio  which  the  first  parallelogram  AB  has  to  the 
other  CD.  Let  this  straight  line 
be  EG;  therefore  the  ratio  of 
AE  to  EG  is  given;  and  EB  is 
to  FD,  as  FC  to  EG,  therefore 
the  ratio  of  FC  to  EG  is  given, 
because  the  ratio  of  EB  to  FD 
is  given;  and  because  the  ratio 
of  AE  to  EG,  as  also  the  ratio 
of  FC  to  EG  is  given;  the  ratio  H  K  L 

of  AE  to  CF  is  given  (9.  dat.). 

The  ratio  of  AE  to  CF  may  be  found  thus:  take  a  straight  line 
H  given  in  magnitude;  and  because  the  ratio  of  the  parallelo- 
gram AB  to  CD  is  given,  make  the  ratio  of  H  to  K  the  same  with 
it.  And  because  the  ratio  of  FD  to  EB  is  given,  make  the  ratio 
of  K  to  L  the  same:  the  ratio  of  AE  to  CF  is  the  same  with  the 
ratio  of  H  to  L.  Make  as  EB  to  FD,  so  FC  to  EG,  therefore,  by 
inversion,  as  FD  to  EB,  so  is  EG  to  FC;  and  as  AE  to  EG,  so  is 
(63.  dat.)  (the  parallelogram  AB  to  CD,  and  so  is)  H  to  K;  but 
as  EG  to  FC,  so  is  (FD  to  EB,  and  so  is)  K  to  L;  therefore  ex 
sequali,  as  AE  to  FC,  so  is  H  to  L. 


EUCLID  S    DATA.  337 


PROP.  LXVI.  69. 

If  two  parallelograms  have  unequal,  but  given  angles, 
and  a  given  ratio  to  one  another;  if  one  side  have 
to  one  side  a  given  ratio,  the  other  side  has  also  a 
ffiven  ratio  to  the  other  side. 

Let  the  two  parallcloj^-rams  ABCD,  EFGH  which  have  the 
given  unequal  angles  ABC,  EFG,  have  a  given  ratio  to  one  ano- 
ther, and  let  the  ratio  of  BC  to  FG  be  given;  the  ratio  also  of  AB 
to  EF  is  given. 

At  the  point  B  of  the  straight  line  BC  make  the  angle  CBK 
equal  to  the  given  angle  EFG,  and  complete  the  parallelogram. 
BKLC;  and  because  each  of  the  angles  BAK,  AKB,  is  given,  the 
triangle  ABK  is  given  (43.  dat.)  in  species;  therefore  the  ratio  of 
AB  to  BK  is  given;  and  because,  by  the  hypothesis,  the  ratio  of 
the  parallelogram  AC  to  EG  is  given,  and  that  AC  is  equal  (35. 
1.)  to  BL;  therefore  the  ratio  of  BL  to  EG  is  given:  and  because 
BL  is  equiangular  to  EG,  and,  by  the  hypothesis,  the  ratio  of  BC 
to  FG  is  given;  therefore  (65.  dat.)  the  ratio  of  KB  to  EF  is  given, 
and  the  ratio  of  KB  to  BA  is  given; 
the  ratio  therefore  (9.  dat.)  of  AB  to 
EF  is  given. 

The  ratio  of  AB  to  EF  may  be  found 

thus:  take  the  straight  line  MN  given      E  . jj      "^^ 

in  position  and  magnitude;  and  make  -r 
the  angle  NMO  equal  to  the  given  an- 
gle BAK,  and  the  angle  MNO  equal 
to  the  given  angle  EFG  or  AKB:  and 
because  the  parallelogram  BL  is  equi- 
angular to  FG,  and  has  a  given  ratio  to  it,  and  that  the  ratio  of 
BC  to  FG  is  given;  find  by  the  65th  dat.  the  ratio  of  KB  to  EF; 
and  make  the  ratio  of  NO  to  OP  the  same  with  it:  then  the  ratio 
of  AB  to  EF  is  the  same  with  the  ratio  of  MO  to  OP:  for  since 
the  triangle  ABK  is  equiangular  to  MON,  as  AB  to  BK,  so  is  MO 
to  ON:  and  as  KB  to  EF,  so  is  NO  to  OP;  therefore,  ex  eequali^ 
as  AB  to  EF,  so  is  MO  to  OP. 

PROP.  LXVn.  70. 

If  the  sides  of  two  equiangular  parallelograms  have 
given  ratios  to  one  another;  the  parallelograms  shall 
have  a  given  ratio  ^o  one  another.^ 

Let  ABCD,  EFGH  be  two  equiangular  parallelograms,  and  let 
the  ratio  of  AB  to  £F,  as  also  the  ratio  of  BC  to  FG,  be  given; 
the  ratio  of  the  parallelogram  AC  to  EG  is  given. 

Take  a  straight  line  K  given  in  magnitude,  and  because  the 

•^  See  Note. 
2  U 


338 


EUCLID  S    DATA. 


ratio  of  A  B  to  EF  is  given,  make 
the  ratio  of  K  to  L  the  same 
with  it;  therefore  L  is  given  (2. 
dat.):  and  because  the  ratio  of 
BC  to  FG  is  given,  make  the 
ratio  of  L  to  M  the  same:  there- 
fore M  is  given  (2.  dat.):  and  Kis 
given,  wherefore  (l.  dat.)  the  ra- 
tio of  K  to  M  is  given:  but  the  parallelogram  AC  is  to  the  paral- 
lelogram EG,  as  the  straight  line  K  to  the  straight  line  M,  as  is 
demonstrated  in  the  23d  prop,  of  B.  6.  Elem.j  therefore  the  ratio 
of  AC  to  EG  is  given. 

From  this  it  is  plain  how  the  ratio  of  two  equiangular  parallelo- 
grams may  be  found  when  the  ratios  of  their  sides  are  given. 


PROP.  LXVIII. 


70. 


If  the  sides  of  two  parallelograms  which  have  un- 
equal, but  given  angles,  have  given  ratios  to  one  ano- 
ther; the  parallelograms  shall  have  a  given  ratio  to 
one  another.* 

Let  two  parallelograms  ABCD,  EFGH,  which  have  the  given 
unequal  angles  ABC,  EFG,  have  the  ratios  of  their  sides,  viz.  of 
AB  to  EF,  and  of  BC  to  FG,  given;  the  ratio  of  the  parallelogram 
AC  to  EG  is  given. 

At  the  point  B  of  the  straight  line  BC  make  the  angle  CBK 
equal  to  the  given  angle  EFG,  and  complete  the  parallelogram 
KBCL;  and  because  each  of  the  angles  BAK,  BKA  is  given,  the 
triangle  ABK  is  given  (43.  dat.)  in  species:  therefore  the  ratio  of 
AB  to  BK  is  given;  and  the  ratio  of  AB  to  EF  is  given,  where- 
fore (9.  dat.)  the  ratio  of  BK  to  K  A  L  D  E  H 
EF  is  given:  and  the  ratio  of 
BC  to  FG  is  given;  and  the  an- 
gle KBC  is  equal  to  the  angle 
EFG;  therefore  (67.  dat.)  the 
ratio  of  the  parallelogram  KC 
to  EG  is  given:  but  KC  is  equal 
(35.  1.)  to  AC;  therefore  the  ra- 
tio of  AC  to  EG  is  given. 

The  ratio  of  the  parallelogram  AC  to  EG  may  be  found  thus: 
take  the  straight  line  MN  given  in  position  and  magnitude,  and 
make  the  angle  MNO  equal  to  the  given  angle  KAB,  and  the 
angle  NMO  equal  to  the  given  angle  AKB,  or  FEH:  and  because 
the  ratio  of  AB  to  EF  is  given,  make  thegratio  of  NO  to  P  the 
same;  also  make  the  ratio  of  P  to  Q  the  same  with  the  given 
ratio  of  BC  to  FG,  the  parallelogram  AG  is  to  EG,  as  MO  to  Q. 

Because  the  angle  KAB  is  equal  to  the  angle  MNO,  and  the 
angle  AKB  equal  to  the  angle  NMO;  the  triangle  AKB  is  equi- 
angular to  NMO:  therefore  as  KB  to  BA,  so  is  MO  to  ON;  and 


■\] 


M 


N 


O 


*  See  Note. 


Euclid's  data. 


339 


G  D 


H 


as  BA  to  EF,  so  is  NO  to  P|  wherefore,  ex  sequali^  as  KB  to  EF, 
so  is  MO  to  P:  and  BC  is  to  FG,  as  P  to  Q,  and  the  parallelo- 
grams KC,  EG  are  equiangular^  therefore,  as  was  shown  in  prop. 

67,  the  parallelog:ram  KC,  that  is,  AC  is  to  EG,  as  MO  to  Q. 
Cor.   1.    If  two  triangles,  ABC,  DEE  have  two  equal  angles, 

or  two  unequal,  but  given  angles,  ABC,  DEF,  and  if  the  ratios 
of  the  sides  about  these  angles,  viz.  the 
ratios  of  AB  to  DE,  and  of  BC  to  EF 
be  given^  the  triangles  shall  have  a  given 
ratio  to  one  another. 

Complete  the  parallelograms  BG,  EH: 
the  ratio  of  BG  to  EH  is  given  (67.  or       B  C        E  F 

68.  dat.)^  and  therefore  the  triangles  which  are  the  halves  (34. 
1.)  of  them  have  a  given  (15.  5.  72.)  ratio  to  one  another. 

Cor.  2.  If  the  bases  BC,  EF  of  two  triangles  ABC,  DEF 
have  a  given  ratio  to  one  another,  and  if  also  the  straight  lines 
AG,  DH  which  are  drawn  to  the  bases  from  the  opposite  angles, 
either  in  equal  angles,  or  unequal,  but  given  angles,  AGC,  DHF 

A 


have  a  given  ratio  to  one  ano 


K 


D 


ther^  the  triangles  shall  have  a 
given  ratio  to  one  another. 

Draw  BK,  EL  parallel  to  AG, 
DH,  and  complete  the  parallelo- 
grams KC,  LF.  And  because 
the  angles  AGC,  DHF,  or  their  equals,  the  angles  KBC,  LEF 
are  either  equal,  or  unequal,  but  given,*  and  that  the  ratio  of  AG 
to  DH,  that  is,  of  KB  to  LE,  is  given,  as  also  the  ratio  of  BC  to 
EF;  therefore  {67.  or  68.  dat.)  the  ratio  of  the  parallelogram 
KC  to  LF  is  given;  wherefore  also  the  ratio  of  the  triangle  ABC 
to  DEF  is  given  (41.  1.  15.  5.). 


PROP.  LXIX. 


61. 


If  a  parallelogram  which  has  a  given  angle  be  applied 
to  one  side  of  a  rectihneal  figure  given  in  species;  if  the 
figure  have  a  given  ratio  to  the  parallelogram,  the  pa- 
rallelogram is  given  in  species. 

Let  ABCD  be  a  rectilineal  figure  given  in  species,  and  to  one 
side  of  it  AB,  let  the  parallelogram  ABEF,  having  the  given 
angle  ABE,  be  applied;  if  the  figure  ABCD  have  a  given  ratio  to 
the  parallelogram  BF,  the  parallelogram  BF  is  given  in  species. 

Through  the  point  A  draw  AG  parallel  to  BC,  and  through 
the  point  C  draw  CG  parallel  to  AB,  and  produce  GA,  CB  to 
the  points  H,  K:  because  the  angle  ABC  is  given  (3.  def.),  and 
the  ratio  of  AB  to  BC  is  given,  the  figure  ABCD  being  given  in 
species;  therefore  the  parallelogram  BG  is  given  (3.  def.)  in 
species.  And  because  upon  the  same  straight  line  AB  the  two 
rectilineal  figures  BD,  BG  given  in  species  are  described,  the  ra- 
tio of  BD  to  BG  is  given  {5c>.  dat.);  and,  by  hypothesis,  the  ratio 
of  BD  to  the  parallelogram  BF  is  given;  wherefore  (9.  dat.)  the 
ratio  of  BF,'  that  is,  (35.   1.)  of  the  parallelogram  BH,  to  BG  is 


340  Euclid's  data. 

given,  and  therefore  (1.  6.)  the  ratio  of  the  straight  line  KB  to 
BC  is  givenj  and  the  ratio  of  BC  to  BA  is  given,  wherefore  the 
ratio  of  KB  to  BA  is  given  (9.  dat.):  and  because  the  angle  ABC 
is  given,  the  adjacent  angle  ABK  is  givenj  and  the  angle  ABE  is 
given,  therefore  the  remaining  angle  KBE  is  given.     The  angle 
EKB  is  also  given,  because  it  is  equal  to  the  angle  ABK;  there- 
fore the  triangle  BKE  is  given  in  species,  and  consequently  the 
ratio  of  EB  to  BK  is  given;  and  the  ratio  of  KB  to  BA  is  givejn, 
wherefore  (9.  dat.)  the  ratio 
of  EB  to  BA  is  given;  and 
the    angle    ABE    is    given, 
therefore  the   parallelogram 
BF  is  given  in  species. 

A  parallelogram  similar  to 
BF  may  be  found  thus:  take 
a  straight  line  LM  given  in 
position  and  magnitude:  and 
because  the  angles  ABK,  ABE  are  given,  make  the  angle  NLM 
equal  to  ABK,  and  the  angle  NLO  equal  to  ABE.  And  because 
the  ratio  of  BF  to  BD  is  given,  make  the  ratio  of  LM  to  P  the 
same  with  it;  and  because  the  ratio  of  the  figure  BD  to  BG  is 
given,  find  this  ratio  by  the  53d  dat.  and  make  the  ratio  of  P  to 
Q  the  same.  Also,  because  the  ratio  of  CB  to  BA  is  given,  make 
the  ratio  of  Q  to  R  the  same;  and  take  LN  equal  to  R;  through 
the  point  M  draw  OM  parallel  to  LN,  and  complete  the  paral- 
lelogram NLOS;  then  this  is  similar  to  the  parallelogram  BF. 

Because  the  angle  ABK  is  equal  to  NLM,  and  the  angle  ABE 
to  NLO,  the  angle  KBE  is  equal  to  MLO;  and  the  angles  BKE, 
LMO  are  equal,  because  the  angle  ABK  is  equal  to  NLM;  there- 
fore the  triangles    BKE,  LMO  are  equiangular  to  one  another; 
wherefore  as  BE   to  BK,  so  is  LO  to  LM:  and  because  as  the 
figure  BF  to  BD,  so  is  the  straight  line  LM  to  P;  and  as  BD  to 
BG,  so  is  P  to  Q;  ex  sequali^  as  BF,  that  is  (35.  1.)  BH  to  BG,  so 
is  LM  to  Q:  but  BH  is  to  (l.  6.)  BG,  as  KB  to  BC:  as  therefore 
KB  to  BC,  so  is  LM  to  Q;  and  because  BE  is  to  BK,  as  LO  to 
LM;  and  as  BK  to  BC,  so  is  LM  to  Q:  and  as  BC  to  BA,  so  Q 
was  made  to  R;  therefore,  ex  aequali,  as  BE  to  BA,  so  is  LO  to 
R,  that  is  to  to  LN;  and  the  angles  ABE,  NLO  are  equal;  there- 
fore the  parallelogram  BF  is  similar  to  LS. 

PROP.  LXX.  62.  78. 

If  two  straight  lines  have  a  given  ratio  to  one 
another,  and  upon  one  of  them  be  described  a  rectih- 
neal  figure  given  in  species,  and  upon  the  other  a 
parallelogram  having  a  given  angle;  if  the  figure  have 
a  given  ratio  to  the  parallelogram,  the  parallelogram  is 
given  in  species.* 

Let  the  two  straight  lines  AB,  CD  have  a  given  ratio  to  one 
another,  and  upon  AB  let  the  figure  AEB  given  in  species  be  de- 

*  See  Note. 


EUCLID  S  DATA.  341 

scribed,  and  upon  CD  the  parallelogram  DF  having  the  given 
angle  FCD;  if  the  ratio  of  AEB  to  DF  be  given,  the  parallelo- 
gram DF  is  given  in  species. 

Upon  the  straight  line  AB,  conceive  the  parallelogram  AG  to 
be  described  similar,  and  similarly  placed  to  FDj  and  because 
the  ratio  of  AB  to  CD  is  given,  and  upon  them  are  described  the 
similar  rectilineal  figures  AG,  FD; 
the  ratio  of  AG  to  FD  is  given  (54. 
dat.);  and  the  ratio  of  FD  to  AEB 
is  given;  therefore  (9.  dat.)  the  ra- 
tio of  AEB  to  AG  is  given;  and  the 
angle  ABG  is  given,  because  it  is 
equal  to  the  angle  FCD:  because 
therefore  the  parallelogram  AG 
which  has  a  given  angle  ABG  is 
applied  to  a  side  AB  of  the  figure 
AEB  given  in  species,  and  the  ratio  of  AEB  to  AG  is  given,  the 
parallelogram  AG  is  given  (69.  dat.)  in  species;  but  FD  is  simi- 
lar to  AG;  therefore  FD  is  given  in  species. 

A  parallelogram  similar  to  FD  may  be  found  thus:  take  a 
straight  line  H  given  in  magnitude;  and  because  the  ratio  of  the 
figure  AEB  to  FD  is  given,  make  the  ratio  of  H  to  K  the  same 
with  it:  also,  because  the  ratio  of  the  straight  line  CD  to  AB  is 
given,  find  by  the  54th  dat.  the  ratio  which  the  figure  FD  de- 
scribed upon  CD  has  to  the  figure  AG  described  upon  AB  simi- 
lar to  FD;  and  make  the  ratio  of  K  to  L  the  same  with  this  ratio; 
and  because  the  ratios  of  H  to  K,  and  of  K  to  L  are  given,  the 
ratio  of  H  to  L  is  given  (9.  dat.);  because,  therefore,  as  AEB  to 
FD,  so  is  H  to  K;  and  as  FD  to  AG,  so  is  K  to  L;  ex  xquali^  as 
AEB  to  AG,  so  is  H  to  L;  therefore  the  ratio  of  AEB  to  AG  is 
given;  and  the  figure  AEB  is  given  in  species,  and  to  its  side  AB 
the  parallelogram  AG  is  applied  in  the  given  angle  ABG;  there- 
fore by  the  69th  dat.  a  parallelogram  may  be  found  similar  to 
AG:  let  this  be  the  parallelogram  MN;  MN  also  is  similar  to 
FD;  for,  by  the  construction,  MN  is  similar  to  AG,  and  AG  is 
similar  to  FD;  therefore  the  parallelogram  FD  is  similar  to 
MN. 

PROP.  LXXI.  81. 

If  the  extremes  of  three  proportional  straight  hnes 
have  given  ratios  to  the  extremes  of  other  three  pro- 
portional straight  lines;  the  means  shall  also  have  a 
given  ratio  to  one  another:  and  if  one  extreme  have 
a  given  ratio  to  one  extreme,  and  the  mean  to  the 
mean;  likewise  the  other  extreme  shall  have  to  the 
other  a  given  ratio. 

Let  A,  B,  C  be  three  proportional  straight  lines,  and  D,  E,  F 
three  other;  and  let  the  ratios  of  A  to  D,  and  of  C  to  F  be  given; 
then  the  ratio  of  B  to  E  is  also  given. 

Because  the  ratio  of  A  to  D,  as  also  of  C  to  F  is  given,  the  ratio 


342 


EUCLID  S    DATA. 


A 

D 
I 


of  the  rectangle  A,  C  to  the  rectangle  D,  F  is  given  (67.  dat.); 
but  the  square  of  B  is  equal  (17.  6.)  to  the  rectangle  A,  C;  and 
the  square  of  E  to  the  rectangle  (17.  6.)  D,  F^  therefore  the  ratio 
of  the  square  of  B  to  the  square  of  E  is  given;  wherefore  (58.  dat.) 
also  the  ratio  of  the  straight  line  B  to  E  is  given. 

Next,  let  the  ratio  of  A  to  D,  and  of  B  to 
E  be  given;  then  the  ratio  of  C  to  F  is  also 
given. 

Because  the  ratio  of  B  to  E  is  given,  the  ratio 
of  the  square  of  B  to  the  square  of  E  is  given 
(54.  dat.);  therefore  (17.  6.)  the  ratio  of  the 
rectangle  A,  C  to  the  rectangle  D,  F  is  given; 
and  the  ratio  of  the  side  A  to  the  side  D  is 
given;  therefore  the  ratio  of  the  other  side  C 
to  the  other  F  is  given  (65.  dat.). 

Cor.  And  if  the  extremes  of  four  proportionals  have  to  the 
extremes  of  four  other  proportionals  given  ratios,  and  one  of  the 
means  a  given  ratio  to  one  of  the  means;  the  other  mean  shall 
have  a  given  ratio  to  the  other  mean,  as  may  be  shovrn  in  the 
same  manner  as  in  the  foregoing  proposition. 


I 
C 


E     F 


PROP.  LXXII. 


82. 


If  four  straight  lines  be  proportionals;  as  the  first  is 
to  the  straight  line  to  which  the  second  has  a  given 
ratio,  so  is  the  third  to  a  straight  line  to  which  the 
fourth  has  a  given  ratio. 

Let  A,  B,  C,  D  be  four  proportional  straight  lines,  viz.  as  A  to 
B,  so  C  to  D;  as  A  is  to  the  straight  line  to  which  B  has  a  given 
ratio,  so  is  C  to  a  straight  line  to  which  D  has  a  given  ratio. 

Let  E  be  the  straight  line  to  which  B  has  a  given 
ratio,  and  as  B  to  E,  so  make  D  to  F:  the  ratio  of 
B  to  E  is  given  (Hyp.),  and  therefore  the  ratio  of  D 
to  F;  and  because  as  A  to  B,  so  is  C  to  D;  and  as  B 
to  E,  so  D  to  F;  therefore,  exsequali,  as  A  to  E,  so  is 
C  to  F;  and  E  is  the  straight  line  to  which  B  has  a 
given  ratio,  and  F  that  to  which  D  has  a  given  ratio; 
therefore  as  A  is  to  the  straight  line  to  which  B  has 
a  given  ratio,  so  is  C  to  a  line  to  which  D  has  a 
given  ratio. 


A 
C 


B 
D 


E 
F 


PROP.  LXXin. 


83. 


If  four  straight  lines  be  proportionals;  as  the  first  is 
to  the  straight  line  to  which  the  second  has  a  given 
ratio,  so  is  a  straight  hne  to  which  the  third  has  a  given 
ratio  to  the  fourth.* 


See  Note, 


EUCLID  S  DATA. 


;43 


A 

F 


B      E 
C     D 


64. 


Let  the  straight  line  A  be  to  B,  as  C  to  D:  as  A 
to  the  straight  line  to  which  B  has  a  given  ratio,  so 
is  a  straight  line  to  which  C  has  a  given  ratio  to  D. 

Let  E  be  the  straight  line  to  which  B  has  a  given 
ratio,  and  as  B  to  E,  so  make  F  to  C;  because  the 
ratio  of  B  to  E  is  given,  the  ratio  of  C  to  F  is  given: 
and  because  A  is  to  B,  as  C  to  Dj  and  as  B  to  E,  so 
F  to  Cj  therefore^ ex  dequaliin proportione perturbata 
(23.  5.))  A  is  to  E,  as  F  to  D;  that  is,  A  is  to  E,  to 
v/hich  B  has  a  given  ratio,  as  F,  to  which  C  has  a 
given  ratio,  is  to  D. 

PROP.  LXXIV. 

If  a  triangle  have  a  given  obtuse  angle;  the  excess 
of  the  square  of  the  side  which  subtends  the  obtuse 
angle,  above  the  squares  of  the  sides  which  contain  it, 
shall  have  a  given  ratio  to  the  triangle. 

Let  the  triangle  ABC  have  a  given  obtuse  angle  ABC;  and 
produce  the  straight  line  CB,  and  from  the  point  A  draw  AD  per- 
pendicular to  BC:  the  excess  of  the  square  of  AC  above  the 
squares  of  AB,  BC,  that  is  (12.  2.),  the  double  of  the  rectangle 
contained  by  DB,  BC,  has  a  given  ratio  to  the  triangle  ABC. 

Because  the  angle  ABC  is  given,  the  angle  ABD  is  also  given; 
and  the  angle  ADB  is  given;  wherefore  the  triangle  ABD  is 
given  (43.  dat.)  in  species;  and  therefore  the  ratio  of  AD  to  DB 
is  given:  and  as  AD  to  DB,  so  is  (l.  6.)  the  rectangle  AD,  BC  to 
the  rectangle  DB,  BC;  wherefore  the  ratio  of  the  rectangle  AD, 
BC  to  the  rectangle  DB,  BC  is  given,  as  also  the  ratio  of  twice 
the  rectangle  DB,  BC  to  the  rectangle  AD,     A  " 

BC:  but  the  ratio  of  the  rectangle  AD,  BC 
to  the  triangle  ABC  is  given,  because  it  is 
double  (41.  1.)  of  the  triangle;  therefore  the 
ratio  of  twice  the  rectangle  DB,  BC  to  the 
triangle  ABC  is  given  (9.  dat.);  and  twice 
the  rectangle  DB,  BC  is  the  excess  (12.  2.)     D     B  C 

of  the  square  of  AC  above  the  squares  of  AB,  BC;  therefore  this 
excess  has  a  given  ratio  to  the  triangle  ABC. 

And  the  ratio  of  this  excess  to  the  triangle  ABC  may  be  found 
thus:  take  a  straight  line  EF  given  in  position  and  magnitude; 
and  because  the  angle  ABC  is  given,  at  the  point  F  of  the  straight 
line  EF,  make  the  angle  EFG  equal  to  the  angle  ABC;  produce 
GF,  and  draw  EH  perpendicular  to  FG;  then  the  ratio  of  the 
excess  of  the  square  of  AC  above  the  squares  of  AB,  BC  to  the 
triangle  ABC,  is  the  same  with  the  ratio  of  quadruple  the  straight 
line  HF  to  HE.  i  r  . 

Because  the  angle  ABD  is  equal  to  the  angle  EFH,  and  the 
angle  ADB  to  EHF,  each  being  a  right  angle:  the  triangle  ADB 
is  equiangular  to  EHF;  therefore  (4.  6.)  as  BD  to  DA,  so  FH  to 
HE;  and  as  quadruple  of  BD  to  DA,  so  is  (cor.  4.  5.)  quadruple 
of  FH  to  HE:  but  as  twice  BD  is  to  DA,  so  is  ( I.  6.)  twice  the 
rectangle  DB.  BC  to  the  rectangle  AD,  BC;  and  as  DA  to  the 


E 


344 


EUCLID  S    DATA. 


half  of  it,  so  is  (cor.  5.)  the  rectangle  AB,  BC  to  its  half  the  tri- 
angle ABC;  therefore,  ex  dsqualij  as  twice  BD  is  to  the  half  of 
DA,  that  is,  as  quadruple  of  BD  is  to  DA,  that  is,  as  quadruple 
of  FH  to  HE,~  so  is  twice  the  rectangle  DB,  BC  to  the  triangle 
ABC. 


PROP.  LXXV. 


65. 


If  a  triangle  have  a  given  acute  angle,  the  space  by 
which  the  square  of  the  side  subtending  the  acute  angle 
is  less  than  the  squares  of  the  sides  which  contain  it, 
shall  have  a  given  ratio  to  the  triangle. 

Let  the  triangle  ABC  have  a  given  acute  angle  ABC,  and  draw 
AD  perpendicular  to  BC;  the  space  by  which  the  square  of  AC 
is  less  than  the  squares  of  AB,  BC,  that  is  (13,  2.),  the  double  of 
the  rectangle  contained  by  CB,  BD,  has  a  given  ratio  to  the  tri- 
angle ABC. 

Because  the  angles  ABD,  ADB  are  each  of  them  given,  the 
triangle  ABD  is  given  in  species;  and  therefore  the  ratio  of  BD  to 
DA  is  given:  and  as  BD  to  DA,  so  is  the  A 

rectangle  CB,  BD  to  the  rectangle  CB,  AD; 
therefore  the  ratio  of  these  rectangles  is 
given,  as  also  the  ratio  of  twice  the  rectan- 
gle CB,  BD  to  the  rectangle  CB,  AD;  but 
the  rectangle  CB,  AD  has  a  given  ratio  to 
its  half  the  triangle  ABC;   therefore  (9.         B  DC 

dat.)  the  ratio  of  twice  the  rectanglejCB,  BD  to  the  triangle  ABC 
is  given;  and  twice  the  rectangle  CB,  BD  is  (13.  2.)  the  space  by 
which  the  square  of  AC  is  less  than  the  squares  of  AB,  BC;  there- 
fore the  ratio  of  this  space  to  the  triangle  ABC  is  given:  and  the 
ratio  may  be  found  as  in  the  preceding  proposition. 


LEMMA. 

If  from  the  vertex  A  of  an  isosceles  triangle  ABC  any  straight 
line  AD  be  drawn  to  the  base  BC,  the  square  of  the  side  AB  is 
equal  to  the  rectangle  BD,  DC  of  the  segments  of  the  base  to- 
gether with  the  square  of  AD;  but  if  AD  be  drawn  to  the  base 
produced,  the  square  of  AD  is  equal  to  the  rectangle  BD,  DC  to- 
gether with  the  square  of  AB. 

Case  1.  Bisect  the  base  BC  in  E,  and  join  A 

AE  which  will  be  perpendicular  (8.  1.)  to 
BC;  wherefore  the  square  of  AB  is  equal  (47. 
1.)  to  the  squares  of  AE,  EB;  but  the  square 
of  EB  is  equal  (5.  1.)  to  the  rectangle  BD, 
DC,  together  with  the  square  of  DE;  there-  t^ 
fore  the  square  of  AB  is  equal  to  the  squares 

of  AE,  ED,  that  is  to  (47.  1.)  the  square  of  AD,  together  with  the 
rectangle  BD,  DC;  the  oth^r  case  is  shown  in  the  same  way  by 
6.  2.  Elem. 


Euclid's  data.  343 

V  PROP.  LXXVI.  67. 

If  a  triangle  have  a  given  angle,  the  excess  of  the 
square  of  the  straight  line  which  is  equal  to  the  two 
sides  that  contain  the  given  angle,  above  the  square  of 
the  third  side,  shall  have  a  given  ratio  to  the  triangle* 

Let  the  triangle  ABC  have  the  given  angle  BAG;  the  excess  of 
the  square  of  the  straight  line  which  is  equal  to  BA,  AC  together 
above  the  square  of  BC,  shall  have  a  given  ratio  to  the  triangle 
ABC. 

Produce  BA,  and  take  AD  equal  to  AC;  join  DC,  and  produce 
it  to  E,  and  through  the  point  B  draw  BE  parallel  to  AC;  join 
AE,  and  draw  AF  perpendicular  to  DC;  and  because  AD  is  equal 
to  AC,  BD  is  equal  to  BE:  and  BC  is  drawn  from  the  vertex  B 
of  the  isosceles  triangle  DBE;  therefore,  by  the  lemma,  the  square 
of  BD,  that  is,  of  BA  and  AC  together,  is  equal  to  the  rectangle 
DC,  CE  together  with  the  square  of  BC;  and  therefore,  the  square 
of  BA,  AC,  together,  that  is,  of  BD:  is  D 

greater  than  the  square  of  BC  by  the 
rectangle  DC,  CE;  and  this  rectangle 
has  a  given  ratio  to  the  triangle  ABC: 

because  the  angle  BAC  is  given,  the  ^^      V-^    C 

adjacent   angle    CAD    is    given:    and  '^ 

each  of  the  angles  ADC,  DCA  is  given, 
for  each  of  them  is  the  half  (5.  8c  32.) 
of  the  given  angle  BAC;  therefore  the 
triangle  ADC  is  given  (43.  dat.)  in 
species;  and  AF  is  drawn  from  its  ver- 
tex to  the  base  in  a  given  angle;  wherefore  the  ratio  of  AF  to  the 
base  CD  is  given  (50.  dat.};  and  as  CD  to  AF,  so  is  (l.  6.)  the 
rectangle  DC,  CE  to  the  rectangle  AF,  CE;  and  the  ratio  of  the 
rectangle  AF,  CE  to  its  half  (41.  1.),  the  triangle  ACE,  is  given; 
therefore  the  ratio  of  the  rectangle  DC,  CE  to  the  triangle  ACE, 
that  is  (37.  1.),  to  the  triangle  ABC,  is  given  (9.  dat.),  and  the 
rectangle  DC,  CE  is  the  excess  of  the  square  of  BA,  AC  together 
above  the  square  of  BC:  therefore  the  ratio  of  this  excess  to  the 
triangle  ABC  is  given. 

The  ratio  which  the  rectangle  DC,CE  has  to  the  triangle  ABC 
is  found  thus:  take  the  straight  line  GH  given  in  position  and 
magnitude,  and  at  the  point  G  in  GH  make  the  angle  HGK  equal 
to  the  given  angle  CAD,  and  take  GK  equal  to  GH;  join  KH,  and 
draw  GL  perpendicular  to  it:  then  the  ratio  of  HK  to  the  half  of 
GL  is  the  same  with  the  ratio  of  the  rectangle  DC,  CE  to  the  tri- 
angle ABC:  because  the  angles  PIGK,  DAC,  at  the  vertices  of  the 
isosceles  triangles  GHK,  ADC  are  equal  to  one  another,  these  tri- 
angles are  similar;  and  because  GL,  AF  are  perpendicular  to  the 
bases  HK,  DC,  as  HK  to  GL,  so  is  (4.  6.  22.  5.)  (DC  to  AF,  and 
so  is)  the  rectangle  DC,  CE  to  the  rectangle  AF,  CE;  but  as  GL 
to  its  half,  so  is  the  rectangle  AF,  CE  to  its  half,  which  is  the  tri- 
angle ACE,  or  the  triangle  ABC;  therefore,  ex  sequali^  HK  is  to 
the  half  of  the  straight  line  GL,  as  the  rectangle  DC,  CE  is  to  the 
triangle  ABC. 
2  X 


546 


EUCLID  S  DATA. 


Cor.  And  if  a  triangle  have  a  given  angle,  the  space  by  which 
the  square  of  the  straight  line  which  is  the  difference  of  the  sides 
which  contain  the  given  angle  is  less  than  the  square  of  the  third 
side,  shall  have  a  given  ratio  to  the  triangle.  This  is  demon- 
strated the  same  way  as  the  preceding  proposition,  by  help  of  the 
second  case  of  the  lemma. 


PROP.  LXXVII. 


L. 


If  the  perpendicular  drawn  from  a  given  angle  of  a 
triangle  to  the  opposite  side,  or  base,  have  a  given 
ratio  to  the  base,  the  triangle  is  given  in  species.^ 

Let  the  triangle  ABC  have  the  given  angle  BAC,  and  let  the 
perpendicular  AD  drawn  to  the  base  BC  have  a  given  ratio  to  it; 
the  triangle  ABC  is  given  in  species. 

If  ABC  be  an  isosceles  triangle,  it  is  evident  (5.  and  32. 1.)  that 


if  any  one  of  its  angles  be  given,  the  rest  are  also  given;  and 
therefore  the  triangle  is  given  in  species,  without  the  considera- 
tion of  the  ratio  of  the  perpendicular  to  the  base,  which  in  this 
case  is  given  by  prop.  50. 

But  when  ABC  is  not  an  isosceles  triangle,  take  any  straight 
line  EF  given  in  position  and  magnitude,  and  upon  it  describe 
the  segment  of  a  circle  EOF  containing  an  angle  equal  to  the 
given  angle  BAC;  draw  GH  bisecting  EF  at  right  angles,  and 
join  EG,  GF:  then,  since  the  angle  EGF  is  equal  to  the  angle 
BAC,  and  that  EGF  is  an  isosceles  triangle,  and  ABC  is  not,  the 
angle  FEG  is  not  equal  to  the  angle  CBA:  draw  EL,  making  the 
angle  FEL  equal  to  the  angle  CBA;  join  FL,  and  draw  LM  per- 
pendicular to  EF;  then,  because  the  triangles  ELF,  BAC  are 
equiangular,  as  also  are  the  triangles  MLE,  DAB,  as  ML  to  LE, 
so  is  DA  to  AB;  and  as  LE  to  EF,  so  is  AB  to  BC;  wherefore, 
ex  dequali,  as  LM  to  EF,  so  is  AD  to  BC;  and  because  the  ratio 
of  AD  to  BC  is  given,  therefore  the  ratio  of  LM  to  EF  is  given; 
and  EF  is  given,  wherefore  (2.  dat.)  LM  also  is  given.  Complete 
the  parallelogram  LMFK;  and  because  LM  is  given,  FK  is  given 
in  magnitude;  it  is  also  given  in  position,  and  the  point  F  is 
given,  and  consequently  (30.  dat.)  the  point  K;  and  because 
through  K  the  straight  line  KL  is  drawn  parallel  to  EF  which  is 
given  in  position,  therefore  (31.  dat.)  KL  is  given  in  position: 
and  the  circumference  ELF  is  given  in  position;  therefore  the 
point  L  is  given  (28.  dat.).    And  because  the  points  L,  E,  F,  are 


See  Note. 


EUCLID  S  DATA. 


347 


given,  the  straight  lines  LE,  EF,  FL,  are  given  (29.  dat.)  in  mag- 
nitude^  therefore  the  triangle  LEF  is  given  in  species  (42.  dat.); 


R  D  C 


and  the  triangle  ABC  is  similar  to  LEF,  wherefore  also  ABC  is 
given  in  species. 

Because  LM  is  less  than  GH,  the  ratio  of  LM  to  EF,  that  is, 
the  given  ratio  of  AD  to  BC,  must  be  less  than  the  ratio  of  GH 
to  EF,  which  the  straight  line,  in  a  segment  of  a  circle  containing 
an  angle  equal  to  the  given  angle,  that  bisects  the  base  of  the  seg- 
ment at  right  angles,  has  unto  the  base. 

Cor.  1.  If  two  triangles,  ABC,  LEF,  have  one  angle  BAC 
equal  to  one  angle  ELF,  and  if  the  perpendicular  AD  be  to  the 
base  BC,  as  the  perpendicular  LM  to  the  base  EF,  the  triangles 
ABC,  LEF  are  similar. 

Describe  the  circle  EFG  about  the  triangle  ELF,  and  drawLN 
parallel  to  EF;  join  EN,  NF,  and  draw  NO  perpendicular  to  EF; 
because  the  angles  ENF,  ELF  are  equal,  and  that  the  angle  EFN 
is  equal  to  the  alternate  angle  FNL,  that  is,  to  the  angle  FEL  in 
the  same  segment;  therefore  the  triangle  NEF  is  similar  to  LEF; 
and  in  the  segment  EGF  there  can  be  no  other  triangle  upon  the 
base  EF,  which  has  the  ratio  of  its  perpendicular  to  that  base  the 
same  with  the  ratio  of  LM  or  NO  to  EF,  because  the  perpendicu- 
lar must  be  greater  or  less  than  LM  or  NO;  but,  as  has  been 
shown  in  the  preceding  demonstration,  a  triangle  similar  to  ABC 
can  be  described  in  the  segment  EGF  upon  the  base  EF,  and  the 
ratio  of  its  perpendicular  to  the  base  is  the  same,  as  was  there 
shown,  with  the  ratio  of  AD  to  BC,  that  is,  of  LM  to  EF;  there- 
fore that  triangle  must  be  either  LEF,  or  NEF,  which  therefore 
are  similar  to  the  triangle  ABC. 

Cor.  2.  If  a  triangle  ABC  have  a  given  angle  BAC,  and  if  the 
straight  line  AR  drawn  from  the  given  angle  to  the  opposite  side 
BC,  in  a  given  angle  ARC,  have  a  given  ratio  to  BC,  the  triangle 
ABC  is  given  in  species. 

Draw  AD  perpendicular  to  BC;  therefore  the  triangle  ARD  is 
given  in  species;  wherefore  the  ratio  of  AD  to  AR  is  given:  and 
the  ratio  of  AR  to  BC  is  given,  and  consequently  (9.  dat.)  the 
ratio  of  AD  to  BC  is  given;  and  the  triangle  ABC  is  therefore 
given  in  species  {J7.  dat.). 

Cor.  3.  If  two  triangles  ABC,  LEF  have  one  angle  BAC  equal 
to  one  angle  ELF,  and  if  straight  lines  drawn  from  these  angles 
to  the  bases,  making  with  them  given  and  equal  angles,  have  the 
same  ratio  to  the  bases,  each  to  each;  then  the  triangles  are  simi- 
lar; for  having  drawn  perpendiculars  to  the  bases  from  the  equal 
angles,  as  one  perpendicular  is  to  its  base,  so  is  the  other  to  its 
base  (4.  6.  22.  5.);  wherefore,  by  Cor.  1.  the  triangles  are  similar. 


348 


Euclid's  data. 


A  triangle  similar  to  ABC  may  be  found  thus:  having  de- 
scribed the  segment  EGF,  and  drawn  the  straight  line  GH,  as 
was  directed  in  the  proposition,  find  FK,  which  has  to  EF  the 
given  ratio  of  AD  to  BC ;  and  place  FK  at  right  angles  to  EF 
from  the  point  F;  then,  because,  as  has  been  shown,  the  ratio  of 
AD  to  BC,  that  is  of  FK  to  EF,  must  be  less  than  the  ratio  of 
GH  to  EF;  therefore  FK  is  less  than  GH;  and  consequently  the 
parallel  to  EF,  drawn  through  the  point  K,  must  meet  the  cir- 
cumference of  the  segment  in  two  points:  let  L  be  either  of  them, 
and  join  EL,  LF,  and  draw  LM  perpendicular  to  EF:  then,  be- 
cause the  angle  BAC  is  equal  to  the  angle  ELF,  and  that  AD  is 
to  BC,  as  KF,  that  is,  LM  to  EF,  the  triangle  ABC  is  similar  to 
the  triangle  LEF,  by  Cor.  1. 


PROP.  LXXVHL 


80. 


If  a  triangle  have  one  angle  given,  and  if  the  ratio 
of  the  rectangle  of  the  sides  which  contain  the  given 
angle  to  the  square  of  the  third  side  be  given,  the  tri- 
angle is  given  in  species. 

Let  the  triangle  ABC  have  the  given  angle  BAC,  and  let  the 
ratio  of  the  rectangle  BA,  AC  to  the  square  of  BC  be  given;  the 
triangle  ABC  is  given  in  species. 

From  the  point  A,  draw  AD  perpendicular  to  BC;  the  rectan- 
gle AD,  BC  has  a  given  ratio  to  its  half  (41.  1.),  the  triangle 
ABC;  and  because  the  angle  BAC  is  given,  the  ratio  of  the  tri- 
angle ABC  to  the  rectangle  BA,  AC  is  given  (Cor.  62.  dat.);  and 
by  the  hypothesis,  the  ratio  of  the  rectangle  BA,  AC  to  the  square 
of  BC  is  given;  therefore  (9.  dat.)  the  ratio  of  the  rectangle  AD, 
BC  to  the  square  of  BC,  that  is  (1.  6.),  the  ratio  of  the  straight 
line  AD  to  BC  is  given;  wherefore  the  triangle  ABC  is  given  in 
species  (77.  dat.). 

A  triangle  similar  to  ABC  may  be  found  thus:  take  a  straight 
line  EF  given  in  position  and  magnitude,  and  make  the  angle 
FEG  equal  to  the  given  angle  BAC,  and  draw  FH  perpendicular 
to  EG,  and  BK  perpendicular  to  AC:  therefore  the  triangle  ABK, 
EFH  are  similar,  and  the 
rectangle  AD,  BC,  or  the 
rectangle  BK,  AC  which  is 
equal  to  it,  is  to  the  rect- 
angle   BA,    AC,     as     the 
straight    line    BK    to    BA, 
that  is,  as  FH  to  FE.     Let 
the  given  ratio  of  the  rect- 
angle BA,  AC  to  the  square  of  BC  be  the  same  with  the  ratio  of 
the  straight  line  EF  to  FL;  therefore,  ex  asqiialij  the  ratio  of  the 
rectangle  AD,  BC  to  the  square  of  BC,  that  is,  the  ratio  of  the 
straight  line  AD  to  BC,  is  the  same  with  the  ratio  of  HF  to  FL; 
and  because  x\D  is  not  greater  than  the  straight  line  MN  in  the 
segment  of  the  circle  described  about  the  triangle  ABC,  which 
bisects  BC  at  right  angles;  the  ratio  of  AD  to  BC,  that  is,  of  HF 


EUCLID  S  DATA.  349 

lo  FL,  must  not  be  greater  than  the  ratio  of  MN  to  BC:  let  it  be 
S05  and,  by  the  77th  clat.  find  a  triangle  OPQ  which  has  one  of 
its  angles  POQ  equal  to  the  given  angle  BAG,  and  the  ratio  of 
the  perpendicular  OR,  drawn  from  that  angle  to  the  base  PQ  the 
same  with  the  ratio  of  HF  to  FL;  then  the  triangle  ABC  is  simi- 
lar to  OPQ:  because,  as  has  been  shown,  the  ratio  of  AD  to  BC  is 
the  same  with  the  ratio  of  (HF  to  FL,  that  is,  by  the  construc- 
tion, with  the  ratio  of)  OR  to  PQ;  and  the  angle  BAC  is  equal 
to  the  angle  POQ;  therefore  the  triangle  ABC  is  similar  (h  Cor. 
77.  dat.)  to  the  triangle  POQ. 

Otherwise^ 

Let  the  triangle  ABC  have  the  given  angle  BAC,  and  let  the 
ratio  of  the  rectangle  BA,  AC,to  the  square  of  BC  be  given;  the 
triangle  ABC  is  given  in  species. 

Because  the  angle  BAC  is  given,  the  excess  of  the  square  of 
both  the  sides  BiV,  AC  together  above  the  square  of  the  third 
side  BC  has  a  given  (76.  dat.)  ratio  to  the  triangle  ABC.  Let  the 
figure  D  be  equal  to  this  excess^  therefore  the  ratio  of  D  to  the 
triangle  ABC  is  given:  and  the  ratio  of  the  triangle  ABC  to  the 
rectangle  BA,  AC  is  given  (Cor.  62.  dat.),  because  BAC  is  a  given 
angle;  and  the  rectangle  BA,  AC  has  a 
given  ratio  to  the  square  of  BC:  where- 
fore (10.  dat.)  the  ratio  of  D  to  the  square 
of  BC  is  given;  and  by  composition  (7. 
dat.)  the  ratio  of  the  space  D  together 
with  the  square  of  BC  to  the  square  of 
BC  is  given;  but  D  together  with  the 
square  of  BC  is  equal  to  the  square  of  both  BA  and  AC  together; 
therefore  the  ratio  of  the  square  of  BA,  AC  together  to  the  square 
of  BC  is  given;  and  the  ratio  of  BA,  AC  together  to  BC  is  there- 
fore given  (59.  dat.);  and  the  angle  BAC  is  given,  wherefore  (48. 
dat.)  the  triangle  ABC  is  given  in  species. 

The  composition  of  this,  which  depends  upon  those  of  the  76th 
and  48th  propositions,  is  more  complex  than  the  preceding  com- 
position, which  depends  upon  that  of  prop.  77,  which  is  easy. 

PROP.  LXXIX.  K. 

If  a  triangle  have  a  given  angle,  and  if  the  straight 
line  drawn  from  that  angle  to  the  base,  making  a  given 
angle  with  it,  divide  the  base  into  segments  which  have 
a  given  ratio  to  one  another;  the  triangle  is  given  in 
species.* 

Let  the  triangle  ABC  have  the  given  angle  BAC,  and  let  the 
straight  line  AD  drawn  to  the  base  BC  making  the  given  angle 
ADB,  divide  BC  into  the  segments  BD,  DC  which  have  a  given 
ratio  to  one  another;  the  triangle  ABC  is  given  in  species. 

Describe  (5.  4.)  the  circle  BAC  about  the  triangle,  and  from  its 

*  See  Note. 


350  Euclid's  DATA. 

centre  E,  draw  EA,  EB,  EC,  EDj  because   the   angle  BAG  is 
given,  the  angle  BEC  at  the  centre,  which  is  the  double  (20.  3.) 
of  it,  is  given.     And  the  ratio  of  BE  to  EC  is  given,  because  they 
are  equal  to  one  another^  therefore  (44.  dat.)  the  triangle  BEC  is 
given  in  species,  and  the  ratio  of  EB  to  BC  is  given^  also  the 
ratio  of  CD  to  BD  is  given  (7.  dat.),  because  the  ratio  of  BD  to 
DC  is  given;  therefore  the  ratio  of  EB  to  BD  is  given  (9.  dat.) 
and  the  angle  EBC  is  given,  wherefore  the  triangle  EBD  is  given 
(9.  dat.)  in  species,  and  the  ratio  of  EB,  that  is,  of  EA  to  ED,  is 
therefore  given;  and  the  angle  EDA  is  given,  because  each  of  the 
angles  BDE,  BDA  is  given;  therefore  the  triangle  AED  is  given 
(47.  dat.)  in  species,  and  the  angle  AED  given; 
also  the  angle  DEC  is  given,  because  each  of 
the  angles  BED,  BEC  is  given;  therefore  the 
angle  AEC  is  given,  and  the  ratio  of  EA  to 
EC,  which  are  equal,  is  given;  and  the  trian- 
gle AEC  is  therefore  given  (44.  dat.)  in  spe-  B 
cies,  and  the  angle  ECA  given;  and  the  angle 
ECB  is  given,  wherefore  the  angle  ACB  is 
given,  and  the  angle  BAC  is  also  given;  therefore  (43.  dat.)  the 
triangle  ABC  is  given  in  species. 

A  triangle  similar  to  ABC  may  be  found,  by  taking  a  straight 
line  given  in  position  and  magnitude,  and  dividing  it  in  the  given 
ratio  which  the  segments  BD,  DC  are  required  to  have  to  one 
another:  then,  if  upon  that  straight  line  a  segment  of  a  circle  be 
described  containing  an  angle  equal  to  the  given  angle  BAC,  and 
a  straight  line  be  drawn  from  the  point  of  division  in  an  angle 
equal  to  the  given  angle  ADB,  and  from  the  point  where  it  meets 
the  circumference,  straight  lines  be  drawn  to  the  extremity  of  the 
first  line,  these,  together  with  the  first  line,  shall  contain  a  triangle 
similar  to  ABC,  as  may  easily  be  shown. 

The  demonstration  may  be  also  made  in  the  manner  of  that  of 
the  77th  prop,  and  that  of  the  77th  may  be  made  in  the  manner 

of  this. 

* 

PROP.  LXXX.  L. 

If  the  sides  about  an  angle  of  a  triangle  have  a  given 
ratio  to  one  another,  and  if  the  perpendicular  drawn 
from  that  angle  to  the  base  have  a  given  ratio  to  the 
base;  the  triangle  is  given  in  species. 

Let  the  sides  BA,  AC,  about  the  angle  BAC  of  the  triangle 
ABC  have  a  given  ratio  to  one  another,  and  let  the  perpendicular 
AD  have  a  given  ratio  to  the  base  BC;  the  triangle  ABC  is  given 
in  species. 

First,  let  the  sides  AB,  AC  be  equal  to  one  another,  therefore 
the  perpendicular  AD  bisects  (26.  1.)  the  base  1  A 
BC;  and  the  ratio  of  AD  to  BC,  and  therefore 
to  its  half  DB,  is  given;  and  the  angle  ADB  is 
given;  wherefore  the  triangle  (43.  dat.)  ABD, 
and  consequently  the  triangle  ABC,  is  given  (44. 
dat.)  in  species. 


Euclid's  data.  351 

But  let  the  sides  be  unequal,  and  BA  be  greater  than  AC;  and 
make  the  angle  CAE,  equal  to  the  angle  ABC;  because  the  angle 
AEB  is  common  to  the  triangles  AEB,  CEA,  they  are  similar; 
therefore  as  AB  to  BE,  so  is  CA  to  AE,  and,  by  permutation,  as 
BA  to  AC,  so  is  BE  to  EA,  and  so  is  EA  to  EC;  and  the  ratio  of 
BA  to  AC  is  given,  therefore  the  ratio  of  BE  to  EA,  and  the  ratio 
of  EA  to  EC,  as  also  the  ratio  of  BE  to  EC  is  given  (9.  dat.); 
wherefore  the  ratio  of  EB  to  BC  is  given  (6.  dat.);  and  the  ratio 
of  AD  to  BC  is  given  by  the  hypothesis,  there-  A 

fore  (9.  dat.)  the  ratio  of  AD  to  BE  is  given; 
and  the  ratio  of  BE  to  EA  was  shown  to  be 
given;  wherefore  the  ratio  of  AD  to  EA  is 
given,  and  ADE  is  a  right  angle,  therefore 
the  triangle  ADE  is  given  (46.  dat.)  in  species, 
and  the  angle  AEB  given;  the  ratio  of  BE  to 
EA  is  likewise  given,  therefore  (44.  dat.)  the  triangle  ABE  is 
given  in  species,  and  consequently  the  angle  EAB,  as  also  the 
angle  ABE,  that  is,  the  angle  CAE,  is  given;  therefore  the  angle 
BAC  is  given,  and  the  angle  ABC  being  also  given,  the  triangle 
ABC  is  given  (43.  dat.)  in  species. 

How  to  find  a  triangle  which  shall  have  the  things  which  arc 
mentioned  to  be  given  in  the  proposition,  is  evident  in  the  first 
case;  and  to  find  it  the  more  easily  in  the  other  case,  it  is  to  be 
observed,  that,  if  the  straight  line  EF  equal  to  EA  be  placed  in 
EB  towards  B,  the  point  F  divides  the  base  BC  into  the  segments 
BF,  FC  which  have  to  one  another  the  ratio  of  the  sides  BA,  AC, 
because  BE,  EA,  or  EF,  and  EC,  were  shown  to  be  proportionals, 
therefore  (19.  5.)  BF  is  to  FC  as  BE  to  EF,  or  EA,  that  is,  as  BA 
to  AC;  and  AE  cannot  be  less  than  the  altitude  of  the  triangle 
ABC,  but  it  may  be  equal  to  it,  which,  if  it  be,  the  triangle,  in  this 
case,  as  also  the  ratio  of  the  sides,  may  be  thus  found:  having 
given  the  ratio  of  the  perpendicular  to  the  base,  take  the  straight 
line  GH  given  in  position  and  magnitude,  for  the  base  of  the  tri- 
angle to  be  found;  and  let  the  given  ratio  of  the  perpendicular  to 
the  base  be  that  of  the  straight  line  K  to  GH,  that  is,  let  K  be 
equal  to  the  perpendicular;  and  suppose  GLH  to  be  the  triangle 
which  is  to  be  found,  therefore  having  made  the  angle  HLM  equal 
to  LGH,  it  is  required  that  LM  be  perpendicular  to  GM,  and 
equal  to  K;  and  because  GM,  ML,  MH  are  proportionals,  as  was 
shown  of  BE,  EA,  EC,  the  rectangle  GMH  is  equal  to  the  square 
of  ML.     Add  the  common  square  of  NH,  (having  bisected  GH  in 
N),  and  the  square  of  NM  is  equal  (6.  2.)  to  the  squares  of  the 
given  straight  lines  NH  and  ML  or  K;  therefore  the  square  of  NM 
and  its  side  NM,  is  given,  as  also  the  point  M,  viz.  by  taking  the 
straight  line  NM,  the  square  of  which  is  equal  to  the  squares  of 
NH,  ML.     Draw  ML  equal  to  K,  at  right  angles  to  GM;  and  be- 
cause ML  is  given  in  position  and  magnitude,  therefore  the  point  L 
is  given;  join  LG,  LH;  then  the  triangle  LGH  is  that  which  was  to 
be  found,  for  the  square  of  NM  is  equal  to  the  squares  of  NH  and 
ML,  and  taking  away  the  common  square  of  NH,  the  rectangle  GMH 


352 


EUCLID  S    DATA. 


is  equal  (6 .2.)  to  the  square  of  ^ 
ML,  therefore  as  GM  to  ML, 
so  is  ML  to  MH,  and  the  tri- 
angle LGM  is  (6.  6.)  therefore 
equiangular  to  HLM,  and  the 
angle  HLM  equal  to  the  an- 
gle LGM,  and  the  straight 
line  LM  drawn  from  the  ver- 
tex of  the  triangle  making  the 
angle  HLM  equal  to  LGH,  is 


S 


N  Q  H 


perpendicular  to  the  base  and  equal  to  the  given  straight  line  K, 
as  was  required;  and  the  ratio  of  the  sides  GL,  LH  is  the  same 
with  the  ratio  of  GM  to  ML,  that  is,  with  the  ratio  of  the  straight 
line  which  is  made  up  of  GN  the  half  of  the  given  base  and  of  NM, 
the  square  of  which  is  equal  to  the  squares  of  GN  and  K,  to  the 
straight  line  K. 

And  whether  this  ratio  of  GM  to  ML  be  greater  or  less 
than  the  ratio  of  the  sides  of  any  other  triangle  upon  the  base  GH, 
and  of  which  the  altitude  is  equal  to  the  straight  line  K,  that  is, 
the  vertex  of  which  is  in  the  parallel  to  GH  drawn  through  the 
point  L,  may  be  thus  found.  Let  OGH  be  any  such  triangle,  and 
draw  OP,  making  the  angle  HOP  equal  to  the  angle  OGH;  there- 
fore, as  before,  GP,  PO,  PH  are  proportionals,  and  PO  cannot  be 
equal  to  I^M,  because  the  rectangle  GPH  would  be  equal  to  the 
rectangle  GMH,  which  is  impossible;  for  the  point  P  cannot  fall 
upon  M,  because  O  would  then  fall  on  L;  nor  can  PO  be  less  than 
LM,  therefore  it  is  greater;  and  consequently  the  rectangle  GPH 
is  greater  than  the  rectangle  GMH,  and  the  straight  line  GP 
greater  than  GM:  therefore,  the  ratio  of  GM  to  MH  is  greater 
than  the  ratio  of  GP  to  PH,  and  the  ratio  of  the  square  of  GM 
to  the  square  of  ML  is  therefore  (2.  cor.  20.  6.)  greater  than  the 
ratio  of  the  square  of  GP  to  the  square  of  PO,  and  the  ratio  of  the 
straight  line  GM  to  ML,  greater  than  the  ratio  of  GP  to  PO. 
But  as  GM  to  ML,  so  is  GL  to  LH;  and  as  GP  to  PO,  so  is  GO 
to  OH;  therefore  the  ratio  of  GL  to  LH  is  greater  than  the  ratio 
of  GO  to  OH;  wherefore  the  ratio  of  GL  to  LH  is  the  greatest  of 
all  others;  and  consequently  the  given  ratio  of  the  greater  side  to 
the  less,  must  not  be  greater  than  this  ratio. 

But  if  the  ratio  of  the  sides  be  not  the  same  with  this  greatest 
ratio  of  GM  to  ML,  it  must  necessarily  be  less  than  it:  let  any  less' 
ratio  be  given,  and  the  same  things  being  supposed,  viz.  that  GH 
is  the  base,  and  K  equal  to  the  altitude  of  the  triangle,  it  may  be 
found  as  follows.  Divide  GH  in  the  point  Q,  so  that  the  ratio  of 
GQ  to  QH  may  be  the  same  with  the  given  ratio  of  the  sides; 
and  as  GQ  to  QH,  so  make  GP  to  PQ,  and  so  will  (19.  5.)  PQ  be 
to  PH;  wherefore  the  square  of  GP  is  to  the  square  of  PQ,  as  (2. 
cor.  29.  6.)  the  straight  line  GP  to  PH:  and  because  GM,  ML, 
MH  are  proportionals,  the  square  of  GM  is  to  the  square  of  ML, 
as  (2.  cor.  20.  6.)  the  straight  line  GM  to  MH:  but  the  ratio  of 
GQ  to  QH,  that  is,  the  ratio  of  GP  to  PQ,  is  less  than  the  ratio  of 
GM  to  ML;  and  therefore  the  ratio  of  the  square  of  GP  to  the 
square  of  PQ  is  less  than  the  ratio  of  the  square  of  GM  to  that  of 


EUCLID  S    DATA.  353 

ML;  and  consequently  the  ratio  of  the  straight  line  GP  to  PH  is 
less  than  the  ratio  of  GM  to  MH;  and  by  division,  the  ratio  of 
GH  to  HP  is  less  than  that  of  GH  to  HM;  wherefore  (10.  5.)  the 
straight  line  HP  is  greater  than  HM,  and  the  rectangle  GPH,  that 
is,  the  square  of  PQ,  greater  than  the  rectangle  GMH,that  is,  than 
the  square  of  ML,  and  the  straight  line  PQ  is  therefore  greater 
than  ML.  Draw  LR  parallel  to  GP,  and  from  P  draw  PR  at 
right  angles  to  GP:  because  PQ  is  greater  than  ML  or  PR,  the 
circle  described  from  the  centre  P,  at  the  distance  PQ,  must  ne- 
cessarily cut  LR  in  two  points;  let  these  be  O,  S  and  join  OG, 
OH;  SG,  SH:  each  of  the  triangles  OGH,  SGH  has  the  things 
mentioned  to  be  given  in  the  proposiiion:  join  OP,  SP;  and  be- 
cause as  GP  to  PQ,  or  PO,  so  is  PO  to  PH,  the  triangle  OGP  is 
equiangular  to  HOP;  as,  therefore,  OG  to  GP,  so  is  HO  to  OP; 
and,  by  permutation,  as  GO  to  OH,  so  is  GP  to  PO,  or  PQ;  and 
so  is  GQ  to  QH:  therefore  the  triangle  OGH  has  the  ratio  of  its 
sides  GO,  OH  the  same  with  the  given  ratio  of  GQ  to  QH:  and  the 
perpendicular  has  to  the  base  the  given  ratio  of  Kto  GH,  because 
the  perpendicular  is  equal  to  LM  or  K:  the  like  may  be  shown  in 
the  same  way  of  the  triangle  SGH. 

This  construction,  by  .which  the  triangle  OGH  is  found  is 
shorter  than  that  which  would  be  deduced  from  the  demonstration 
of  the  datum,  by  reason  that  the  base  GH  is  given  in  position  and 
magnitude,  which  was  not  supposed  in  the  demonstration;  the 
same  thing  is  to  be  observed  in  the  next  proposition. 

PROP.  LXXXI.  M. 

If  the  sides  about  an  angle  of  a  triangle  be  unequal, 
and  have  a  given  ratio  to  one  another,  and  if  the  per- 
pendicular from  that  angle  to  the  base  divide  it  into 
segments  that  have  a  given  ratio  to  one  another,  the 
triangle  is  given  in  species. 

Let  ABC  be  a  triangle,  the  sides  of  which  about  the  angle  BAG 
are  unequal,  and  have  a  given  ratio  to  one  another,  and  let  the  per- 
pendicular AD  to  the  base  BC  divide  it  into  the  segments  BD, 
DC,  which  have  a  given  ratio  to  one  another,  the  triangle  ABC  is 
given  in  species. 

Let  AB  be  greater  than  AC,  and  make  the  angle  CAE  equal  to 
the  angle  ABC;  and  because  the  angle  AEB  is  common  to  the 
triangles  ABE,  CAE,  they  are  (4.  6.)  equian-  A 

gular  to  one  another:  therefore  as  AB  to  BE,  ^/^IV^^ 

so  is  CA  to  AE,  and  by  permutation,  as  AB  to         y^     \\  ^v 
AC,  so  BE  to  EA,  and  so  is  EA  to  EC:  but     B  DC     E 

the  ratio  of  BA  to  AC  is  given,  therefore  the  ratio  of  BE  to  EA,  as 
also  the  ratio  of  EA  to  EC  is  given;  where- 
fore (9.  dat.)  the  ratio  of  BE  to  EC,  as  also 
(cor.  6.  dat.)  the  ratio  of  EC  to  CB  is  given: 
and  the  ratio  of  BC  to  CD  is  given  (7.  dat.) 
because  the  ratio  of  BD  to  DC  is  given; 
therefore  (9.  dat.)  the  ratio  of  EC  to  CD  is 
given, and  consequently  (7.dat.)  the  ratio  of 
2  Y 


354  Euclid's  data. 

DE  to  EC:  and  the  ratio  of  EC  to  EA  was  shown  to  be  given,  there- 
fore (9.  dat.)  the  ratio  of  DE  to  EA  is  g-iven;  and  ADE  is  a  right 
angle,  wherefore  (46.  dat.)  the  triangle  ADE  is  given  in  species,  and 
the  angle  AED  given:  and  the  ratio  of  CE  to  EA  is  given,  there- 
fore (44.  dat.)  the  triangle  AEC  is  given  in  species,  and  conse- 
quently the  angle  ACE  is  given,  as  also  the  adjacent  angle  ACB, 
In  the  same  manner,  because  the  ratio  of  BE  to  EA  is  given,  the 
triangle  BEA  is  given  in  species,  and  the  angle  ABE  is  therefore 
given:  and  the  angle  ACB  is  given;  wherefore  the  triangle  ABC 
is  given  (43.  dat.)  in  species. 

But  the  ratio  of  the  greater  side  BA  to  the  other  AC  must  be 
less  than  the  ratio  of  the  greater  segment  BD  to  DC:  because  the 
square  of  BA  is  to  the  square  of  AC,  as  the  squares  of  BD,  DA 
to  the  squares  of  DC,  DA;  and  the  squares  of  BD,  DA  have  to 
the  squares  of  DC,  DA  a  less  ratio  than  the  square  of  BD  has  to 
the  square  of  DC,*  because  the  square  of  BD  is  greater  than  the 
square  of  DC;  therefore  the  square  of  BA  has  to  the  square  of 
AC  a  less  ratio  than  the  square  of  BD  has  to  that  of  DC:  and 
consequently  the  ratio  of  BA  to  x\C  is  less  than  the  ratio  of  BD 
to  DC. 

This  being  premised,  a  triangle  which  shall  have  the  things 
mentioned  to  be  given  in  the  proposition,  and  to  which  the  trian- 
gle ABC  is  similar,  may  be  found  thus:  take  a  straight  line  GH 
given  in  position  and  magnitude,  and  divide  it  in  K,  so  that  the 
ratio  of  GK  to  KH  may  be  the  same  with  the  given  ratio  of  BA 
to  AC:  divide  also  GH  in  L,  so  that  the  ratio  of  GL  to  LH  may 
be  the  same  with  the  given  ratio  of  BD  to  DC,  and  draw  LM  at 
right  angles  to  GH:  and  because  the  ratio  of  the  side  of  a  trian- 
gle is  less  than  the  ratio  of  the  segments  of  the  base,  as  has  been 
shown,  the  ratio  of  GK  to  KH  is  less  than  the  ratio  of  GL  to  LH; 
wherefore  the  point  L  must  fall  between  K  and  H:  also  make  as 
GK  to  KH;  so  GN  to  NK,  and  so  shall  (19.  5.)  NK  be  to  NH. 
And  from  the  centre  N,  at  the  distance  NK,  describe  a  circle, 
and  let  its  circumference  meet  LM  in  O,  and  join  OG,  OH;  then 
OGH  is  the  triangle  which  was  to  be  described:  because  GN  is 
to  NK,  or  NO,  as  NO  to  NH,  the  triangle  OGN  is  equiangular 
to  HON;  therefore  as  OG  to  GN,  so  is  HO  to  ON,  and  by  per- 
mutation, as  GO  to  OH,  so  is  GN  to  NO,  or  NK,  that  is,  as  GK 
to  KH,  that  is,  in  the  given  ratio  of  the  sides;  and  by  the  con- 
struction, GL,  LH  have  to  one  another  the  given  ratio  of  the  seg- 
ments of  the  base. 

PROP.  LXXXH.  60. 

If  a  parallelogram  given  in  species  and  magnitude 
be  increased  or  diminished  by  a  gnomon  given  in  mag- 

*  If  A  be  greater  than  B,  and  C  any  third  magnitude ;  then  A  and  C  together 
have  to  B  and  C  together  a  less  ratio  than  A  has  to  B. 

Let  A  be  to  B  as  C  to  D,  and  because  A  is  greater  than  B,  C  is  greater  than 
D :  but  as  A  is  to  B,  so  A  and  C  to  B  and  D ;  and  A  and  C  have  to  B  and  C  a 
less  ratio  than  A  and  C  have  to  B  and  D,  because  C  is  greater  than  D,  therefore 
A  and  C  have  to  B  and  C  a  less  ratio  than  A  to  B. 


- 

13 

\ 

F   D        A 


K 


Euclid's  data.  355 

nitude,  the  sides  of  the  gnomon  are  given  in  magni- 
tude. 

First,  let  the  parallelogram  AB  given  in  species  and  magnitude 
be  increased  by  the  given  gnomon  ECBDFG,  each  of  the  straight 
lines  CE,  DF  is  given. 

Because  AB  is  given  in  species  and  magnitude,  and  that  the 
gnomon  ECBDFG  is  given,  therefore  the  whole  space  AG  is 
given  in  magnitude:  but  AG  is  also  given  in  species,  because  it 
is  similar  (2.  def.  2.  and  24.  6.)  to  AB;  therefore  the  sides  of  AG 
are  given  (60.  dat.):  each  of  the  straight  lines      Gl^^xTF  1  E. 

AE,  AF   is   therefore  given;   and  each  of  the  ^  <;;^] C 

straight  lines  CA,  AD  is  given  (60.  dat.),  there- 
fore each  of  the  remainders  EC,  DF  is  given 
(4.  dat.). 

Next,  let  the  parallelogram  AG  given  in  spe- 
cies and  magnitude,  be  diminished  by  the  given 
gnomon  ECBDFG,  each  of  the  straight  lines 
CE,  DF  is  given. 

Because  the  parallelogram  AG  is  given,  as  also  its  gnomon 
ECBDFG,  the  remaining  space  AB  is  given  in  magnitude:  but 
it  is  also  given  in  species:  because  it  is  similar  (2.  def.  2.  and  24. 
6.)  to  AG;  therefore  (60.  dat.)  its  sides  CA,  AD  are  given,  and 
each  of  the  straight  lines  EA,  AF  is  given;  therefore  EC,  DF  are 
each  of  them  given. 

The  gnomon  and  its  sides  CE,  DF  may  be  found  thus  in  the 
first  case.  Let  H  be  the  given  space  to  which  the  gnomon  must 
be  made  equal,  and  find  (25.  6.)  a  parallelogram  similar  to  AB 
and  equal  to  the  figures  AB  and  H  together,  and  place  its  sides 
AE,  AF  from  the  point  A,  upon  the  straight  lines  AC,  AD,  and 
complete  the  parallelogram  AG  which  is  about  the  same  diame- 
ter (26.  6.)  with  AB;  because  therefore  AG  is  equal  to  both  AB 
and  H,  take  away  the  common  part  AB,  the  remaining  gnomon 
ECBDFG  is  equal  to  the  remaining  figure  H;  therefore  a  gno- 
mon equal  to  H,  and  its  sides  CE,  DF  are  found:  and  in  like 
manner  they  may  be  found  in  the  other  case,  in  which  the  given 
figure  H  must  be  less  than  the  figure  FE  from  which  it  is  to  be 
taken. 

PROP.  LXXXIII.  58. 

If  a  parallelogram  equal  to  a  given  space  be  applied 
to  a  given  straight  line,  deficient  by  a  parallelogram 
given  in  species,  the  sides  of  the  defect  are  given. 

Let  the  parallelogram  AC  equal  to  a  given  space  be  applied 
to  the  given  straight  line  AB,  deficient  by  the  parallelogram 
BDCL  given  in  species;  each  of  the  straight  lines  CD,  DB  are 
given. 

Bisect  AB  in  E;  therefore  EB  is  given  in  magnitude:  upon 
EB  describe  (18.  6.)  the  parallelogram  EF  similar  to  DL  and  si- 


356  Euclid's  data. 

milarly  placed;  therefore  EF  is  given  in  spe- 
cies, and  is  about  the  same  diameter  (26.  6.) 
with  DL;  let  BCG  be  the  diameter,  and 
construct  the  figure;  therefoie,  because  the 
figure  EF  given  in  species  is  described  upon 
the  given  straight  line  EB,  EF  is  given  (56. 
dat.)  in  magnitude,  and  the  gnomon  ELH  is 
equal  (36.  and  43.  1.)  to  the  given  figure  AC:  therefore  (82.  dat.) 
since  EF  is  diminished  by  the  given  gnomon  ELH,  the  sides  EK, 
FH  of  the  gnomon  are  given;  but  EK  is  equal  to  DC,  and  FH  to 
DB;  wherefore  CD,  DB  are  each  of  them  given. 

This  demonstration  is  the  analysis  of  the  problem  in  the  28th 
prop,  of  book  6,  the  construction  and  demonstration  of  which 
proposition  is  the  composition  of  the  analysis;  and  because  the 
given  space  AC  or  its  equal  the  gnomon  ELH  is  to  be  taken  from 
the  figure  EF  described  upon  the  half  of  AB  similar  to  BC,  there- 
fore AC  must  not  be  greater  than  EF,  as  is  shown  in  the  27th 
prop.  B,  6. 

PROP.  LXXXIV.  59. 

If  a  parallelogram  equal  to  a  given  space  be  applied 
to  a  given  straight  line,  exceeding  by  a  parallelogram 
given  in  species;  the  sides  of  the  excess  are  given. 

Let  the  parallelogram  AC  equal  to  a  given  space  be  applied 
to  the  given  straight  line  AB,  exceeding  by  the  parallelogram 
BDCL  given  in  species;  each  of  the  straight  lines  CD,  DB  are 
given. 

Bisect  AB  in  E;  therefore  EB  is  given  in  magnitude:  upon 
EB  describe  (18.  6.)  the  parallelogram  EF  similar  to  LD,  and  si- 
milarly placed;  therefore  EF  is  given  in  species,  and  is  about  the 
same  diameter  (26.  6.)  with  LD.    Let  CBG  G       F     H 

be  the  diameter,  and  construct  the  figure: 
therefore,  because  the  figure  EF  given  in  "E 

species  is  described  upon  the  given  straight     Af~ ^hr — '  D 

line  EB,  EF  is  given  in  magnitude,  (56.         j^  ^ 

dat.)  and  the  gnomon  ELH  is  equal  to  the  j^        L"   C 

given  figure  (36.  dat.  43.  1.)  AC;  where- 
fore, since  EF  is  increased  by  the  given  gnomon  ELH,  its  sides 
EK,  FH  are  given  (82.  dat.);  but  EK  is  equal  to  CD,  and  FH  to 
BD;  therefore  CD,  DB  are  each  of  them  given. 

This  demonstration  is  the  analysis  of  the  problem  in  the  29th 
prop,  book  6,  the  construction  and  demonstration  of  which  is  the 
composition  of  the  analysis. 

Cor.  If  a  parallelogram  given  in  species  be  applied  to  a  given 
straight  line,  exceeding  by  a  parallelogram  equal  to  a  given  space; 
the  sides  of  the  parallelogram  are  given. 

Let  the  parallelogram  ADCE  given  in  species  be  applied  to 
the  given  straight  line  AB,  exceeding  by  the  parallelogram  BDCG 
equal  to  a  given  space;  the  sides  AD,  DC  of  the  parallelogram 
are  given. 

Draw  the  diameter  DE  of  the  parallelogram  AC,  and  construct 


B 


1 


X 

r 

i::x>) 

Euclid's  data.  357 

the  figure.  Because  the  parallelogram  AK  is  equal  (43.  l.)to  BC 
which   is    given,  therefore   AK  is    given;         E  G       C 

and  BK  is  similar  (24.  6.)  to  AC,  therefore 
BK  is  given  in  species.  And  since  the 
parallelogram  AK  given  in  magnitude  is 

applied  to  the  given  straight  line  AB,  ex-     H ^ix.    "^K 

ceeding  by  the  parallelogram  BK  given  in 
species,  therefore,  by  this  proposition  BD,         A  B      D 

DK  the  sides  of  the  excess  are  given,  and  the  straight  line  AB  is 
given^  therefore  the  whole  AD,  as  also  DC,  to  which  it  has  a 
given  ratio,  is  given. 

PROBLEM. 

To  apply  a  parallelogram  similar  to  a  given  one  to  a  given 
straight  line  AB,  exceeding  by  a  parallelogram  equal  to  a  given 
space. 

To  the  given  straight  line  AB  apply  (29.  6.)  the  parallelogram 
AK  equal  to  the  given  space,  exceeding  by  the  parallelogram  BK 
similar  to  the  one  given.  Draw  DF,  the  diameter  of  BK,  and 
through  the  point  A  draw  AE  parallel  to  BF,  meeting  DF  pro- 
duced in  E,  and  complete  the  parallelogram  AC. 

The  parallelogram  BC  is  equal  (43.  1.)  to  AK,  that  is,  to  the 
given  space;  and  the  parallelogram  AC  is  similar  (24.  6.)  to  BK; 
therefore  the  parallelogram  AC  is  applied  to  the  straight  line  AB 
similar  to  the  one  given,  and  exceeding  by  the  parallelogram  BC 
which  is  equal  to  the  given  space. 

PROP.  LXXXV.  84. 

If  two  straight  lines  contain  a  parallelogram  given 
in  magnitude,  in  a  given  angle ;  if  the  difference  of  the 
straight  lines  be  given,  they  shall  each  of  them  be 
given. 

Let  AB,  BC  contain  the  parallelogram  AC  given  in  magnitude, 
in  the  given  angle  ABC,  and  let  the  excess  of  BC  above  AB  be 
given;  each  of  the  straight  lines  AB,  BC  is  given. 

Let  DC  be  the  given  excess  of  BC  above  BA,        A  E 

therefore  the  remainder  BD    is  equal    to  BA.  i  T 

Complete  the  parallelogram  AD;  and  because       /  / 


7 


AB  is  equal  to  BD,  the  ratio  of  AB  to  BD  is      /  / 

given;  and  the   angle  ABD  is  given,  therefore     ^ tt 

the  parallelogram  AD  is  given  in  species;  and 
because  the  given  parallelogram  AC  is  applied  to  the  given 
straight  line  DC,  exceeding  by  the  parallelogram  AD  given  in 
species,  the  sides  of  the  excess  are  given  (84.  dat.):  therefore  BD 
is  given;  and  DC  is  given,  wherefore  the  whole  BC  is  given:  and 
AB  is  given,  therefore  AB,  BC  are  each  of  them  given. 

PROP.  LXXXVI.  85. 

If  two  straight  lines  contain  a  parallelogram  given 
in  magnitude,  in  a  given  angle;  if  both  of  them  together 
be  given,  they  shall  each  of  them  be  given. 


358  Euclid's  dat^. 

Let  the  two  straight  lines  AB,  BC  contain  the  parallelogram 
AC  given  in  magnitude,  in  the  given  angle  ABC,  and  let  AB,  BC 
together  be  given^  each  of  the  straight  lines  AB,  BC  is  given.  ' 

Produce  CB,  and  make  BD  equal  to  AB,  and  complete  the 
parallelogram  ABDE.  Because  DB  is  equal  to  BA,  and  the  angle 
ABD  given,  because  the  adjacent  angle  ABC  is         j^  ^ 

given,  the  parallelogram  AD  is  given  in  spe-  y  y -j 

cies;  and  because  AB,  BC  together  are  given,         /  /      / 

and  AB  is  equal  to  BD;  therefore  DC  is  given;       /  I       I 

and  because  the  given  parallelogram  AC  is  ap-      '  f ' 

plied  to  the  given  straight  line  DC,  deficient  by     D  B      C 

the  parallelogram  AD  given  in  species,  the  sides  AB,  BD  of  the 
defect  are  given  (83.  dat.);  and  DC  is  given,  wherefore  the  re- 
mainder BC  is  given;  and  each  of  the  straight  lines  AB,  BC  is 
therefore  given. 

PROP.  LXXXVII.  87. 

If  two  straight  lines  contain  a  parallelogram  given 
in  magnitude,  in  a  given  angle;  if  the  excess  of  the 
square  of  the  greater  above  the  square  of  the  lesser  be 
given,  each  of  the  straight  lines  shall  be  given. 

Let  the  two  straight  lines  AB,  BC  contain  the  given  parallelo- 
gram AC  in  the  given  angle  ABC;  if  the  excess  of  the  square  of 
BC  above  the  square  of  BA  be  given,  AB  and  BC  are  each  of  them 
given. 

Let  the  given  excess  of  the  square  of  BC  above  the  square  of 
BA  be  the  rectangle  CB,  BD:  take  this  from  the  square  of  BC; 
the  remainder,  which  is  (2.  2.)  the  rectangle  BC,  CD,  is  equal  to 
the  square  of  AB;  and  because  the  angle  ABC  of  the  parallelo- 
gram AC  is  given,  the  ratio  of  the  rectangle  of  the  sides  AB,  BC 
to  the  parallelogram  AC  is  given  (62.  dat.);  and  AC  is  given, 
therefore  the  rectangle  AB,  BC  is  given;  and  the  rectangle  CB 
BD  is  given;  therefore  the  ratio  of  the  rectangle  CB,  BD  to  the 
rectangle  AB,  BC,  that  is  (l.  6.),  the  ratio  of  the  straight  line  DB 
to  BA  is  given;  therefore  (54.  dat.)  the  ratio  of  the  square  of  DB 
to  the  square  of  BA  is  given:  and  the  square  of 
BA  is  equal  to  the  rectangle  BC,  CD:  where- 
fore the  ratio  of  the  rectangle  BC,  CD  to  the 
square  of  BD  is  given,  as  also  the  ratio  of  four 


times  the  rectangle  BC,  CD  to  the  square  of     D~pr\ 

BD;  and,  by  composition  (7.  dat.),  the  ratio  of 
four  times  the  rectangle  BC,  CD,  together  with  the  square  of  BC 
to  the  square  of  BD  is  given:  but  four  times  the  rectangle  BC, 
CD,  together  with  the  square  of  BD  is  equal  (8.  2.)  to  the  square 
of  the  straight  lines  BC,  CD  taken  together:  therefore  the  ratio 
of  the  square  of  BC,  CD  together  to  the  square  of  BD  is  given; 
wherefore  (58.  dat.)  the  ratio  of  the  straight  line  BC,  together 
with  CD  to  BD,  is  given:  and,  by  composition,  the  ratio  of  BC, 
together  with  CD  and  DB,  that  is,  the  ratio  of  twice  BC  to  BD, 
is  given;  therefore  the  ratio  of  BC  to  BD  is  given,  as  also  (1.  6.) 
the  ratio  of  the  square  of  BC  to  the  rectangle  CB,  BD:  but  the 


Euclid's  data.  359 

rectangle  CB,  BD  is  given,  being  the  given  excess  of  the  square 
of  BCj  BA5  therefore  the  square  of  BC,  and  the  straight  line  BC, 
is  given:  and  the  ratio  of  BC  to  BD,  as  also  of  BD  to  BA,  has 
been  shown  to  be  given;  therefore  (9.  dat.)  the  ratio  of  BC  to  BA 
is  given;  and  BC  is  given,  wherefore  BA  is  given. 

The  preceding  demonstration  is  the  analysis  of  this  problem,  viz. 

A  parallelogram  AC  which  has  a  given  angle  ABC  being  given 
in  magnitude,  and  the  excess  of  the  square  of  BC  one  of  its  sides 
above  the  square  of  the  other  BA  being  given;  to  find  the  sides: 
and  the  composition  is  as  follows. 

Ket  EFG  be  the  given  angle  to  which  the  angle  ABC  is  re- 
quired to  be  equal,  and  from  any  point  E  in  EF,  draw  EG  per- 
pendicular to  FG;  let  the  rectangle  EG,         -^ 
GH  be  the   given  space  to  which   the 
parallelogram  AC  is  to  be  made  equal; 
and  the  rectangle  HG,  GL  be  the  given 
excess  of  the  squares  of  BC,  BA. 

Take,  in   the  straight  line   GE,   GK 
equal  to  FE,  and  make  GM  double  of 

GK:  join  ML,  and  in  GL  produced,  take     ^ — i J — ^  ^^ 

LN  equal  to  LM:  bisect  GN  in  O,  and     ^     ^       ^    ^  "^ 

between  GH,  GO  find  a  mean  proportional  BC:  as  OG  to  GL,  so 
make  CB  to  BD;  and  make  the  angle  CBA  equal  to  GFE,  and 
as  LG  to  GK,  so  make  DB  to  BA;  and  complete  the  parallelo- 
gram AC:  AC  is  equal  to  the  rectangle  EG,  GH,  and  the  excess 
of  the  squares  of  CB,  BA  is  equal  to  the  rectangle  HG,  GL. 

Because  as  CB  to  BD,  so  is  OG  to  GL,  the  square  of  CB  is  to 
the  rectangle  CB,  BD  as  (l.  6.)  the  rectangle  HG,  GO  to  the 
rectangle  HG,  GL:  and  the  square  of  CB  is  equal  to  the  rectan- 
gle HG,  GO,  because  GO,  BC,  GH  are  proportionals;  therefore 
the  rectangle  CB,  BD  is  equal  (14.  5.)  to  HG,  GL.  And  because 
as  CB  to  BD,  so  is  OG  to  GL;  twice  CB  is  to  BD,  as  twice  OG, 
that  is  GN,  to  GL:  and,  by  division,  as  BC  together  with  CD  is 
to  BD,  so  is  NL,  that  is,LM,  to  LG:  therefore  (22.  6.)  the  square 
of  BC  together  with  CD  is  to  the  square  of  BD,  as  the  square  of 
ML  to  the  square  of  LG:  but  the  square  of  BC  and  CD  together 
is  equal  (8.  2.)  to  four  times  the  rectangle  BC,  CD  together  with 
the  square  of  BD;  therefore  four  times  the  rectangle  BC,  CD 
together  with  the  square  of  BD  is  to  the  square  of  BD,  as  the 
square  of  ML  to  the  square  of  LG:  and,  by  division,  four  times 
the  rectangle  BC,  CD  is  to  the  square  of  BD,  as  the  square  of 
MG  to  the  square  of  GL;  wherefore  the  rectangle  BC,  CD  is  to 
the  square  of  BD  as  (the  square  of  KG  the  half  of  MG  to  the 
square  of  GL,  that  is,  as)  the  square  of  AB  to  the  square  of  BD, 
because  as  LG  to  GK,  so  DB  was  made  to  BA:  therefore  (14.  5.) 
the  rectangle  BC,  CD  is  equal  to  the  square  of  AB.  To  each  of 
these  add  the  rectangle  CB,  BD,  and  the  square  of  BC  becomes 
equal  to  the  square  of  AB  together  with  the  rectangle  CB,  BD; 
therefore  this  rectangle,  that  is,  the  given  rectangle  KG,  GL,  is 
the  excess  of  the  squares  of  BC,  AB.  From  the  point  A,  draw 
AP  perpendicular  to  BC,  and  because  the  angle  ABP  is  equal  to 
the  angle  EFG,  the  triangle  ABP  is  equiangular  to  EFG:  and 


360  Euclid's  data. 

DB  was  made  to  BA,  as  LG  to  GK;  therefore  as  the  rectangle 
CB,  BD  to  CB,  BA,  so  is  the  rectangle  HG,  GL  to  HG,  GK^ 
and  as   the  rectangle   CB,  BA  to  AP,  BC,  so  is  (the   straight 

M 


K 

E/ 
/ 


\ 


z 


F  G        L     O         HN 

line  BA  to  AP,  and  so  is  FE  or  GK  to  EG,  and  so  is)  the  rect- 
angle HG,  GK  to  HG,  GE;  therefore  ex  seqiiali^  as  the  rect- 
angle CB,  BD  to  AP,  BC,  so  is  the  rectangle  HG,  GL  to  EG, 
GH:  and  the  rectangle  CB,  BD  is  equal  to  HG,  GL:  therefore 
the  rectangle  AP,  BC,  that  is,  the  parallelogram  AC,  is  equal  to 
the  given  rectangle  EG,  GH. 

PROP.  LXXXVHL  N. 

If  two  straight  lines  contain  a  parallelogram  given  in 
magnitude,  in  a  given  angle;  if  the  sum  of  the  squares 
of  its  sides  be  given,  the  sides  shall  each  of  them  be 
given. 

Let  the  two  straight  lines  AB,  BC  contain  the  parallelogram 
ABCD  given  in  magnitude  in  the  given  angle  ABC,  and  let  the 
sum  of  the  squares  of  AB,  BC  be  given;  AB,  BC  are  each  of 
them  given. 

First,  let  ABC  be  a  right  angle;  and  because  twice  the  rect- 
angle contained  by  two  equal  straight  lines  is  equal  to  both  their 
squares;  but  if  two  straight  lines  are  unequal. 


twice  the  rectangle  contained  by  them  is  less  A  D 

than  the  sum  of  their  squares,  as  is  evident  from  3  q 

the  7th  prop,  book  2,  Elem.;  therefore  twice  the  

given  space,  to  which  space  the  rectangle  of  which  the  sides  are 
to  be  found  is  equal,  must  not  be  greater  than  the  given  sum  of 
the  squares  of  the  sides:  and  if  twice  that  space  be  equal  to  the 
given  sum  of  the  squares,  the  sides  of  the  rectangle  must  neces- 
sarily be  equal  to  one  another:  therefore  in  this  case  describe  a 
square  ABCD  equal  to  the  given  rectangle,  and  its  sides  AB, 
BC  are  those  which  were  to  be  found:  for  the  rectangle  AC  is 
equal  to  the  given  space,  and  the  sum  of  the  squares  of  its  sides 
AB,  BC  is  equal  to  twice  the  rectangle  AC,  that  is,  by  the  hy- 
pothesis, to  the  given  space  to  which  the  sum  of  the  squares  was 
required  to  be  equal. 

But  if  twice  the  given  rectangle  be  not  equal  to  the  given  sum 
of  the  squares  of  the  sides,  it  must  be  less  than  it,  as  has  been 
shown.  Let  ABCD  be  the  rectangle;  join  AC  and  draw  BE 
perpendicular  to  it,  and  complete  the  rectangle  AEBF,  and  de- 
scribe the  circle  ABC  about  the  triangle  ABC;  AC  is  its  diame- 
ter (Cor.  5.  4.):  and  because  the  triangle  ABC  is  similar  (8.  6.) 
to  AEB,  as  AC  to  CB,  so  is  AB  to  BE;  therefore  the  rectangle 


Euclid's  data.  361 

AC,  BE  is  equal  to  AB,  BCj  and  the  rectangle  AB,  BC  is  given, 
wherefore  AC,  BE  is  given:  and  because  the  sum  of  the  squares 
of  AB,  BC  is  given,  the  square  of  AC  which  is  equal  (47.  1.)  to 
that  sum  is  given;  and  AC  itself  is  therefore  given  in  magnitude: 
let  AC  be  likewise  given  in  position,  and  the  point  A;  therefore 
AF  is  given  (32.  dat.)  in  position :  and  the  A  D 

rectangle  AC,  BE  is  given,  as  has  been 
shown,  and  AC  is  given,  wherefore  (61. 
dat.)  BE  is  given  in  magnitude,  as  also 
AF  which  is  equal  to  it;  and  AF  is  also 
given  in  position,  and  the  point  A  is 
given;  wherefore  (30.  dat.)  the  point  F 
is  given,  and  the  straight  line  FB  in 
position  (31.  dat.):  and  the  circumfe- 
rence ABC  is  given  in  position,  where- 
fore (28.  dat.)  the  point  B  is  given:  and  the  points  A,  C  are  given; 
therefore  the  straight  lines  AB,  BC  are  given  (29.  dat.)  in  posi- 
tion and  magnitude. 

The  sides  AB,  BC  of  the  rectangle  may  b?,  found  thus:  let  the 
rectangle  GH,  GK  be  the  given  space  to  which  the  i^ectangle 
AB,  BC  is  equal;  and  let  GH,  GL  be  the  given  rectangle  to 
which  the  sum  of  the  squares  of  AB,  BC  is  equal:  find  (14.  2.) 
a  square  equal  to  the  rectangle  GH,  GL:  and  let  its  side  AC  be 
given  in  position:  upon  AC  as  a  diameter  describe  the  semicircle 
ABC,  and  as  AC  to  GH,  so  make  GK  to  AF,  and  from  the  point 
A  place  AF  at  right  angles  to  AC:  therefore  the  rectangle  CA, 
AF  is  equal  (16.  6.)  to  GH,  GK;  and,  by  the  hypothesis,  twice 
the  rectangle  GH,  GK  is  less  than  GH,  GL,  that  is,  than  the 
square  of  AC;  wherefore  twice  the  rectangle  CA,  AF  is  less  than 
the  square  of  AC,  and  the  rectangle  CA,  AF  itself  less  than 
half  the  square  of  AC,  that  is,  than  the  rectangle  contained  by 
the  diameter  AC  and  its  half;  wherefore  AF  is  less  than  the  semi- 
diameter  of  the  circle,  and  consequently  the  straight  line  drawn 
through  the  point  F  parallel  to  AC  must  meet  the  circumference 
in  two  points:  let  B  be  either  of  them,  and  join  AB,  BC,  and  com- 
plete the  rectangle  ABCD;  ABCD  is  the  rectangle  which  was  to 
be  found:  draw  BE  perpendicular  to  AC;  therefore  BE  is  equal 
(34.  1.)  to  AF  and  because  the  angle  ABC  in  a  semicircle  is  a 
right  angle,  the  rectangle  AB,  BC  is  equal  (8.  6.)  to  AC,  BE, 
that  is,  to  the  rectangle  CA,  AF,  which  is  equal  to  the  given 
rectangle  GH,  GK:  and  the  squares  of  AB,  BC  are  together  equal 
(47.  1.)  to  the  square  of  AC,  that  is,  to  the  given  rectangle  GH, 
GL. 

But  if  the  given  angle  ABC  of  the  parallelogram  AC  be  not 
a  right  angle,  in  this  case,  because  ABC  is  a  given  angle,  the 
ratio  of  the  rectangle  contained  by  the  sides  AB,  BC  to  the  pa- 
rallelogram AC  is  given  (62.  dat.);  and  AC  is  given,  therefore 
the  rectangle  AB,  BC  is  given;  and  the  sum  of  the  square  of  AB, 
BC  is  given;  therefore  the  sides  AB,  BC  are  given  by  the  pre- 
ceding case. 

The  sides  AB,   BC  and  the  parallelogram   AC  may  be  found 
thus:  let  EFG  be  the  given  angle  of  the  parallelogram,  and  from 
2  Z 


362 


EUCLID  S  DATA. 


any  point  E  in  FE  draw  EG  perpendicular  to  FG;  and  let  the 
rectangle  EG,  FH  be  the  given  space  to  which  the  parallelogram 
is  to  be  made  equal,  and   let  EF,  FK  be   the  '  ^ 

given  rectangle  to  which  the  sum  of  the  square 
of  the  sides  is  to  be  equal.  And,  by  the  pre- 
ceding case,  find  the  sides  of  a  rectangle 
which  is  equal  to  the  given  rectangle  EF,  FH, 
and  the  squares  of  the  sides  of  which  are  to- 
gether equal  to  the  given  rectangle  EF,  FK; 
therefore,  as  was  shown  in  that  case,  twice  the 
rectangle  EF,  FH  must  not  be  greater  than  the 
rectangle  EF,  FK;  let  it  be  so,  and  let  AB, 
BC  be  the  sides  of  the  rectangle  joined  in  the 
angle  ABC  equal  to  the  given  angle  EFG,  and 
complete   the   parallelogram    ABCD,    which        F  H  G     K 

will  be  that  which  was  to  be  found:  draw  AL  perpendicular  to 
BC,  and  because  the  angle  ABL  is  equal  to  EFG,  the  triangle 
ABL  is  equiangular  to  EFG;  and  the  parallelogram  AC,  that  is, 
the  rectangle  AL,  BC  is  to  the  rectangle  AB,  BC,  as  (the  straight 
line  AL  to  AB,  that  is,  as  EG  to  EF,  that  is,  as)  the  rectangle 
EG,  FH,  to  EF,  FH:  and,  by  the  construction,  the  rectangle  AB, 
BC  is  equal  to  EF,  FH,  therefore  the  rectangle  AL,  BC,  or,  its 
equal,  the  parallelogram  AC,  is  equal  to  the  given  rectangle  EG, 
FH;  and  the  squares  of  AB,  BC  are  together  equal,  by  construc- 
tion, to  the  given  rectangle  EF,  FK, 

PROP.  LXXXIX.  86. 

If  two  Straight  lines  contain  a  given  parallelogram 
in  a  given  angle,  and  if  the  excess  of  the  square  of  one 
of  them  above  a  given  space,  has  a  given  ratio  to  the 
square  of  the  other;  each  of  the  straight  lines  shall  be 
given. 

Let  the  two  straight  lines  AB,  BC  contain  the  given  parallelo- 
gram AC  in  the  given  angle  ABC,  and  let  the  excess  of  the  square 
of  BC  above  a  given  space  have  a  given  ratio  to  the  square  of 
AB,  each  of  the  straight  lines  AB,  BC  is  given. 

Because  the  excess  of  the  square  of  BC  above  a  given  space 
has  a  given  ratio  to  the  square  of  BA,  let  the  rectangle  CB,  BD 
be  the  given  space;  take  this  from  the  square  of  BC;  the  remain- 
der, to  wit,  the  rectangle  (2.  2.)  BC,  CB  has  a  given  ratio  to  the 
square  of  BA:  draw  AE  perpendicular  to  BC,  and  let  jLhe  square 
of  BF  be  equal  to  the  rectangle  BC,  CD;  then,  because  the  angle 
ABC,  as  also  BEA,  is  given,  the  triangle  ABE  ^ 

is  given  (43.  dat.)  in  species,  and  the  ratio  of 

AE  to  AB  is  given:  and  because  the  ratio  of     Js^L- - 

the  rectangle  BC,  CD,  that  is,  of  the  square         /j  / 

of  BF  to  the  square  of  BA  is  given,  the  ratio       /!  / 

of  the  straight  line  BF  to  BA  is  given  (58.     /-J ' 

dat.);  and  the  ratio  of  AE  to  AB  is  given,     BED         C 
wherefore  (9.  dat.)  the  ratio  of  AE  to  BF  is  given;  as  also  the 
ratio  of  the  rectangle  AE  to   BC,  that  is,  (35.  1.)  of  the  paral- 


Euclid's  data.  363 

lelogram  AC  to  the  rectangle  FB,  BC;  and  AC  is  given,  where- 
fore the  rectangle  FB,  BC  is  given.  The  excess  of  the  square  of 
BC  above  the  square  of  BF,  that  is,  above  the  rectangle  BC, 
CD  is  given,  for  it  is  equal  (3.  2.)  to  the  given  rectangle  CB,  BDj 
therefore,  because  the  rectangle  contained  by  the  straight  lines 
FB,  BC  is  given,  and  also  the  excess  of  the  square  of  BC  above 
the  square  of  BF;  FB,  BC  are  each  of  them  given  (87.  dat.); 
and  the  ratio  of  FB  to  BA  is  givenj  therefore,  AB,  BC  are  given. 

The  composition  is  as  follows : 

Let  GHK  be  the  given  angle  to  which  the  angle  of  the  paral- 
lelogram is  to  be  made  equal,  and  from  any  point  G  in  HG,  draw 
GK  perpendicular  to  HK^  let  GK,  HL  be  the  rectangle  to  which 
the  parallelogram  is  to  be  made  equal,  and  let 
LH,  HM  be  the  rectangle  equal  to  the  given 
space  which  is  to  be  taken  from  the  square  of 
one  of  the  sidesj  and  let  the  ratio  of  the  re- 
mainder to  the  square  of  the  other  side  be  the  

same  with  the  ratio  of  the  square  of  the  given     H  KM  L 

straight  line  NH  to  the  square  of  the  given  straight  line  HG. 

By  help  of  the  87th  dat.  find  two  straight  lines  BC,  BF,  which 
contain  a  rectangle  equal  to  the  given  rectangle  NH,  HL,  and  such 
that  the  excess  of  the  square  of  BC  above  the  F 

square  of  BF  be  equal  to  the   given  rectangle  / 

LH,  HM;  and  join  CB,  BF  in 'the  angle  FBC    AZ 


equal  to  the  given  angle  GHK:  and  as  NH   to      /  / 

HG,  so  make  FB  to  BA,  and  complete  the  pa-  /   I    .  / 

rallelogram  AC,  and  draw  AE  perpendicular  to  B  E  D  C 

BC;  then  AC  is  equal  to  the  rectangle  GK,  AL;  and  if  from  the 
square  of  BC,  the  given  rectangle  LH,  HM  be  taken,  the  re- 
mainder shall  have  to  the  square  of  BA  the  same  ratio  which  the 
square  of  NH  has  to  the  square  of  HG. 

Because,  by  the  construction,  the  square  of  BC  is  equal  to  the 
square  of  BF,  together  with  the  rectangle  LH,  HM;  if  from  the 
square  of  BC  there  be  taken  the  rectangle  LH,  HM,  there  re- 
miains  the  square  of  BF,  w  hich  has  (22.  6.)  to  the  square  of  BA 
the  same  ratio  which  the  square  of  NH  has  to  the  square  of  HG, 
because,  as  NH  to  HG,  so  FB  was  made  to  BA;  but  as  HG  to 
GK,  so  is  BA  to  AE,  because  the  triangle  GHK  is  equiangular 
to  ABE;  therefore,  ex  eequali,  as  NH  to  GK,  so  is  FB  to  AE; 
wherefore  (l.  6.)  the  rectangle  NH,  HL  is  to  the  rectangle  GK, 
HL,  as  the  rectangle  FB,  BC  to  AE,  BC;  but  by  the  construc- 
tion, the  rectangle  NH,  HL  is  equal  toFB,  BC;  therefore  (14.  5.) 
the  rectangle  GK,  HL  is  equal  to  the  rectangle  AE,  BC,  that  is, 
to  the  parallelogram  AC. 

The  analysis  of  this  problem  might  have  been  made  as  in  the 
86th  prop,  in  the  Greek,  and  the  composition  of  it  may  be  made 
as  that  which  is  in  prop.  87th  of  this  edition. 

PROP.  XC.  O. 

If  two  straight  lines  contain  a  given  parallelogram 
in  a  given  angle,  and  if  the  square  of  one  of  them 


364  Euclid's  data. 

together  with  the  space  which  has  a  given  ratio  to  the 
square  of  the  other  be  given,  each  of  the  straight  Knes 
shall  be  given. 

Let  the  two  straight  lines  AB,  BC  contain  the  given  parallel- 
ogram AC  in  the  given  angle  ABC,  and  let  the  square  of  BC  to- 
gether vi^ith  the  space  v/hich  has  a  given  ratio  to  the  square  of 
AB  be  given;  AB,  BC  are  each  of  them  given. 

Let  the  square  of  BD  be  the  space  which  has  the  given  ratia,' 
to  the  square  of  AB;  therefore,  by  the  hypothesis,  the  square  of 
BC  together  with  the  square  of  BD  is  given.  From  the  point  A, 
draw  AE  perpendicular  to  BC;  and  because  the  angles  ABE, 
BEA  are  given,  the  triangle  ABE  is  given  (43.  dat.)  in  species; 
therefore  the  ratio  of  BA  to  AE  is  given;  and  because  the  ratio 
of  the  square  of  BD  to  the  scjuare  of  BA  is  given,  the  ratio  of 
the  straight  line  BD  to  BA  is  given  (58.  dat.);  and  the  ratio  of 
BA  to  AE  is  given;  therefore  (9.  dat.)  the  ratio  of  AE  to  BD  is 
given,  as  also  the  ratio  of  the  rectangle  AE,  BC,  that  is,  of  the 
parallelogram  AC  to  the  rectangle  DB,  BC;  and  AC  is  given, 
therefore  the  rectangle  DB,  BC  is  given;  and  the  square  of  BC 

D  M 


7 


BE  C 


G  H  K 


together  with  the  square  of  BD  is  given;  therefore  (88.  dat.)  be- 
cause the  rectangle  contained  by  two  straight  lines  DB,  BC  is 
given,  and  the  sum  of  their  squares  is  given,  the  straight  line 
DB,  BC  are  each  of  them  given;  and  the  ratio  of  DB  to  BA  is 
given:  therefore  AB,  BC  are  given. 

The  composition  is  as  follows : 

Let  FGH  be  the  given  angle  to  which  the  angle  of  the  parallel- 
ogram is  to  be  made  equal,  and  from  any  point  F  in  GF  draw 
FH  perpendicular  to  GH;  and  let  the  rectangle  FH,  GK  be  that 
to  which  the  parallelogram  is  to  be  made  equal;  and  let  the  rect- 
angle KG,  GL  be  the  space  to  which  the  square  of  one  of  the  sides 
of  the  parallelogram  together  with  the  space  which  has  a  given 
ratio  to  the  square  of  the  other  side,  is  to  be  made  equal;  and  let 
this  given  ratio  be  the  same  which  the  square  of  the  given  straight 
line  MG  has  to  the  square  of  GF, 

By  the  88th  dat.  find  two  straight  lines  DB,  BC  which  contain 
a  rectangle  equal  to  the  given  rectangle  MG,  GK:  and  such  that 
the  sum  of  their  squares  is  equal  to  the  given  rectangle  KG,  GL: 
therefore,  by  the  determination  of  the  problem  in  that  proposi- 
tion, twice  the  rectangle  MG,  GK,  must  not  be  greater  than  the 
rectangle  KG,  GL.  Let  it  be  so,  and  join  the  straight  lines  DB, 
BC  in  the  angle  DBC  equal  to  the  given  angle  FGH;  and,  as  MG 
to  GF,  so  make  DB  to  BA,  and  complete  the  parallelogram  AC: 
AC  is  equal  to  the  rectangle  FH,  GK;  and  the  square  of  BC  to- 


EUCLID  S  DATA. 


365 


D 

a/ 


7 


B  E 


G  H 


K 


gether  with  the  square  of  BD,  which,  by  the  construction,  has  to 
the  square  of  BA,  the  given  ratio  which  the  square  of  MG  has  to 
the  square  of  GF,  is  equal,  by  the  construction,  to  the  given  rect- 
angle KG,  GL.     Draw  AE  perpendicular  to  BC. 

Because,  as  DB  to  BA,  so  is  MG  to  GFj  and  as  BA  to  AE,  so 
GF  to  FH;  ex  eeqiiali^  as  DB  to  AE,  so  is  MG  to  FH;  therefore 
as  the  rectangle  DB,  BC  to  AE,  BC,  so  is  the  rectangle  MG,  GK 
to  FH,  GKj  and  the  rectangle  DB,  BC  is  equal  to  the  rectangle 
MG,  GK;  therefore  the  rectangle  AE,  BC,  that  is,  the  parallelo- 
gram AC,  is  equal  to  the  rectangle  FH,  GK. 

PROP.  XCI.  88. 

If  a  straight  line  drawn  within  a  circle  given  in  mag- 
nitude cuts  off  a  segment  which  contains  a  given  angle ; 
the  straight  line  is  given  in  magnitude. 

In  the  circle  ABC  given  in  magnitude,  let  the  straight  line  AC 
be  drawn,  cutting  off  the  segment  AEC  which  contains  the  given 
angle  AEC;  the  straight  line  AC  is  given  in  magnitude. 

Take  D  the  centre  of  the  circle  (l.  3.),  join  AD  and  produce  it 
to  E,  and  join  EC:  the  angle  ACE  being  a 
right  (31.  3.)  angle,  is  given;  and  the  angle 
AEC  is  given;  therefore  (43.  dat.)  the  tri- 
angle ACE  is  given  in  species,  and  the  ra- 
tio of  EA  to  AC  is  therefore  given;  and  EA 
is  given  in  magnitude,  because  the  circle  is 
given  (5.  def.)  in  magnitude;  AC  is  there- 
fore given  (2.  dat.)  in  magnitude. 

PROP.  XCII.  89 

a 

If  a  straight  line  given  in  magnitude  be  drawn  within 
a  circle  given  in  magnitude,  it  shall  cut  off  a  segment 
containing  a  given  angle. 

Let  the  straight  line  AC  given  in  magnitude  be  drawn  within 
the  circle  ABC  given  in  magnitude;  it  shall  cut  off  a  segment  con- 
taining a  given  angle. 

Take  D  the  centre  of  the  circle,  join  AD 
and  produce  it  to  E,  and  join  EC:  and  be- 
cause each  of  the  straight  lines  EA  and  AC 
is  given,  their  ratio  is  given  (1.  dat.);  and 
the  angle  ACE  is  a  right  angle,  therefore 
the  triangle  ACE  is  given  (46.  dat.)  in  spe- 
cies, and  consequently  the  angle  AEC  is 
given. 


B 


366 


EUCLID  S    DATA. 


PROP.    XCIII.  90. 

If  from  any  point  in  the  circumference  of  a  circle 
given  in  position,  two  straight  hnes  be  drawn  meeting 
the  circumference,  and  containing  a  given  angle;  if  the 
point  in  which  one  of  them  meets  the  circumference 
again  be  given,  the  point  in  which  the  other  meets  it  is 
also  given. 

From  any  point  A  in  the  circumference  of  a  circle  ABC  given 
in  position,  let  AB,  AC  be  drawn  to  the  cir- 
cumference, miaking  the  given  angle  BAC; 
if  the  point  B  be  given,  the  point  C  is  also 
given. 

Take  D  the  centre  of  the  circle,  and  join 
BD,  DC;   and   because  each  of  the  points  B 
B,  D  is  given,  BD  is  given  (29.  dat.)  in  posi- 
tion; and  because  the  angle  BAC  is  given, 
the  angle  BDC  is  given  (20.  3.),  therefore  be- 
cause the  straight  line  DC  is  drawn  to  the  given  point  D  in  the 
straight  line  BD  given  in  position  in  the  given  angle  BDC,  DC  is 
given  (32.  dat.)  in  position:  and  the  circumference  ABC  is  given 
in  position,  therefore  (28.  dat.)  the  point  C  is  given. 

PROP.  XCIV.  91. 

If  from  a  given  point  a  straight  line  be  drawn  touch- 
ing a  circle  given  in  position ;  the  straight  line  is  given 
in  position  and  magnitude. 

Let  the  straight  line  AB  be  drawn  from  the  given  point  A 
touching  the  circle  BC  given  in  position;  AB  is  given  in  position 
and  magnitude.  fe 

Take  D  the  centre  of  the  circle,  and  join  DA,  DB:  because 
each  of  the  points  D,  A  is  given,  the 
straight  line  AD  is  given  (29.  dat.)  in 
position  and  magnitude;  and  DBA  is  a 
right  (18.  3.)  angle,  wherefore  DA  is  a 
diameter  (Cor.  5.  4.)  of  the  circle  DBA, 
described  about  the  triangle  DBA;  and 
that  circle  is  therefore  given  (6.  def.)  in 
position:  and  the  circle  BC  is  given  in 
position,  therefore  the  point  B  is  given  (28.  dat,);  the  point  A  is 
also  given:  therefore  the  straight  line  AB  is  given  (29.  dat.)  in 
position  and  magnitude. 

PROP.  XCV.  92. 

If  a  straight  line  be  drawn  from  a  given  point  with- 
out a  circle  given  in  position;  the  rectangle  contained 
by  the  segments  betwixt  the  point  and  the  circumfer- 
ence of  the  circle  is  given. 


EUCLID  S  DATA. 


367 


15  A. 


Let  the  straight  line  ABC  be  drawn  from  the  given  point  A 
without  the  circle  BCD  given  in  posi- 
tion, cutting  it  in  B,  C^  the  rectangle 
BA,  AC  is  given. 

From  the  point  A  draw  (17.  3.)  AD      C 
touching  the  circle;  therefore  AD   is 
given  (94.  dat.)  in  position  and  magni- 
tude;  and  because  AD   is   given,  the 
square  of  AD  is  given  (56.  dat.),  which 

is  equal  (36.  8.)  to  the  rectangle  BA,  AC:  therefore  the  rectangle 
BA,  AC  is  given. 

PROP.  XCVI.  93. 

If  a  straight  line  be  drawn  through  a  given  point 
within  a  circle  given  in  position,  the  rectangle  contained 
by  the  segments  betwixt  the  point  and  the  circumfer- 
ence of  the  circle  is  given. 

Let  the  straight  line  BAC  be  drawn  through  the  given  point  A 
within  the  circle  BCE  given  in  position;  the  rectangle  BA,  AC  is 
given. 

Take  D  the  centre  of  the  circle,  join  AD, 
and  produce  it  to  the  points  E,  F;  because 
the  points  A,  D  are  given,  the  straight  line 
AD  is  given  (29.  dat.)  in  position;  and  the 
circle  BEC  is  given  in  position;  therefore 
the  points  E,  F  are  given  (28.  dat.);  and  the 
point  E  is  given,  therefore  EA,  AF  are  each 
of  them  given  (29.  dat.);  and  the  rectangle 
EA,  AF  is  therefore  given;  and  it  is  equal  (35.  3.)  to  the  rectangle 
BA,  AC,  which  consequently  is  given. 


PROP.  xcvn. 


94. 


If  a  straight  line  be  drawn  within  a  circle  given  in 
magnitude,  cutting  off  a  segment  containing  a  given 
angle;  if  the  angle  in  the  segment  be  bisected  by  a 
straight  line  produced  till  it  meets  the  circumference, 
the  straight  lines  which  contain  the  given  angle  shall 
both  of  them  together  have  a  given  ratio  to  the  straight 
line  which  bisects  the  angle:  and  the  rectangle  con- 
tained by  both  these  lines  together  which  contain  the 
given  angle,  and  the  part  of  the  bisecting  line  cut  oft* 
below  the  base  of  the  segment,  shall  be  given. 

Let  the  straight  line  BC  be  drawn  within  the  circle  ABC  given 
in  magnitude,  cutting  off  a  segment  containing  the  given  angle 
BAC,  and  let  the  angle  BAC  be  bisected  by  the  straight  line 
AD;  BA  together  with  AC  has  a  given  ratio  to  AD;  and  the 
rectangle  contained  by  BA  and  AC  together,  and  the  straight 


368  Euclid's  data. 

line  ED  cut  off  from  AD  below  BC  the  base  of  the  segment,  is 
given. 

Join  BD;  and  because  BC  is  drawn  within  the  circle  ABC 
given  in  magnitude,  cutting  off  the  segment  BAC,  containing  the 
given  angle  BAC;  BC  is  given  (91.  dat.)  in  magnitude:  by  the 
same  reason  BD  is  given:  therefore  (l.  dat.)  the  ratio  of  BC  to 
BD  is  given:  and  because  the  angle  BAC  is  bisected  by  AD,  as 
BA  to  AC,  so  is  (3.  6.)  BE  to  EG;  and,  by  permutation,  as  AB  to 
BE,  so  is  AC  to  CE;  wherefore  (12.  5.)  as  BA  and  AC  together 
to  BC,  so  is  AC  to  CE:  and  because  the  angle  BAE  is  equal  to 
EAC,  and  the  angle  ACE  to  (21.  3.)  F 
ABD,  the  triangle  ACE  is  equiangular 
to  the  triangle  ADB;  therefore  as  AC 
to  CE,  so  is  AD  to  DB:  but  as  AC  to 
CE,  so  is  BA  together  with  AC  to  BC: 
as  therefore  BA  and  AC  to  BC,  so  is 
AD  to  DB;  and,  by  permutation,  as  BA 
and  AC  to  AD,  so  is  BC  to  BD:  and  the 
ratio  of  BC  to  BD  is  given,  therefore  the 
ratio  of  BA  together  with  AC  to  AD  is 
given. 

Also  the  rectangle  contained  by  BA  and  AC  together,  and  DE 
is  given. 

Because  the  triangle  BDE  is  equiangular  to  the  triangle  ACE, 
as  BD  to  DE,  so  is  AC  to  CE;  and  as  AC  to  CE,  so  is  BA  and 
AC  to  BC;  therefore  as  BA  and  AC  to  BC,  so  is  BD  to  DE; 
wherefore  the  rectangle  contained  by  BA  and  AC  together,  and 
DE,  is  equal  to  the  rectangle  CB,  BD:  but  CB,  BD  is  given; 
therefore  the  rectangle  contained  by  BA  and  AC  together,  and 
DE,  is  given. 

Othenoise^ 

Produce  CA,  and  make  AF  equal  to  AB,  and  join  BF;  and  be- 
cause the  angle  BAC  is  double  (5.  and  32.  1.)  of  each  of  the  an- 
gles BFA,  BAD,  the  angle  BFA  is  equal  to  BAD;  and  the  angle 
BCA  is  equal  to  BDA,  therefore  the  triangle  FCB  is  equiangular 
to  ABD:  as  therefore  FC  to  CB,  so  is  AD  to  DB;  and,  by  per- 
mutation, as  FC,  that  is,  BA  and  AC  together,  to  AD,  so  is  CB 
to  BD:  and  the  ratio  of  CB  to  BD  is  given,  therefore  the  ratio  of 
BA  and  AC  to  AD  is  given. 

And  because  the  angle  BFC  is  equal  to  the  angle  DAC,  that  is, 
to  the  angle  DEC,  and  the  angle  ACB  equal  to  the  angle  ADB, 
the  triangle  FCB  is  equiangular  to  BDE;  as  therefore  FC  to  CB, 
so  is  BD  to  DE;  therefore  the  rectangle  contained  by  FC,  that  is, 
BA  and  AC  together,  and  DE,  is  equal  to  the  rectangle  CB,  BD, 
which  is  given,  and  therefore  the  rectangle  contained  by  BA,  AC 
together,  and  DE,  is  given. 

PROP.  XCVIII.  P. 

If  a  straight  line  be  drawn  within  a  circle  given  in 
magnitude,  cutting  off  a  segment  containing  a  given 


Euclid's  data.  369 

angle:  if  the  angle  adjacent  to  the  angle  in  the  seg- 
ment be  bisected  by  a  straight  line  produced  till  it  meet 
the  circumference  again  and  the  base  of  the  segment ; 
the  excess  of  the  straight  lines  which  contain  the  given 
angle  shall  have  a  given  ratio  to  the  segment  of  the 
bisecting  line  which  is  within  the  circle;  and  the  rect- 
angle contained  by  the  same  excess  and  the  segment 
of  the  bisecting  line  betwixt  the  base  produced  and  the 
point  where  it  again  meets  the  circumference,  shall  be 
given. 

Let  the  straig-ht  line  BC  be  drawn  within  the  circle  ABC  given 
in  magnitude,  cutting  off*  a  segment  containing  the  given  angle 
BAG,  and  let  the  angle  CAF  adjacent  to  BAG  be  bisected  by  the 
straight  line  DAE,  meeting  the  circumference  again  in  D,  and 
BG  the  base  of  the  segment  produced  in  E;  the  excess  of  BA, 
AG  has  a  given  ratio  to  AD;  and  the  rectangle  which  is  con- 
tained by  the  same  excess  and  the  straight  line  ED,  is  given. 

Join  BD,  and  through  B  draw  BG  parallel  to  DE  meeting  AC 
produced  in  G:  and  because  BG  cuts  off"  from  the  circle  ABC 
given   in   magnitude   the    segment  ~  ~ 

BAG  containing  a  given  angle,  BG 
is  therefore  given  (91.  dat.)  in  mag- 
nitude: bv  the  same  reason  BD  is 
given,  because  the  angle  BAD  is 
equal  to  the  given  angle  CAF: 
therefore  the  ratio  of  BG  to  BD  is 
given:  and  because  the  angle  GAE 
is  equal  to  EAF,  of  which  GAE  is 
equal  to  the  alternate  angle  AGB, 
and  EAF  to  the  interior  and  opposite  angle  ABG;  therefore  the 
angle  AGB  is  equal  to  ABG,  and  the  straight  line  AB  equal  to 
AG;  so  that  GG  is  the  excess  of  BA,  AG;  and  because  the  angle 
BGC  is  equal  to  GAE,  that  is,  to  EAF,  or  the  angle  BAD;  and 
that  the  angle  BGG  is  equal  to  the  opposite  interior  angle  BDA 
of  the  quadrilateral  BGAD  in  the  circle;  therefore  the  triangle 
BGG  is  equiangular  to  BDA:  therefore  as  GG  to  GB,  so  is  AD 
to  DB;  and,  by  permutation,  as  GG  which  is  the  excess  of  BA, 
AG  to  AD,  so  is  GB  to  BD:  and  the  ratio  of  GB  to  BD  is  given: 
therefore  the  ratio  of  the  excess  of  BA,  AG  to  AD  is  given. 

And  because  the  angle  GBG  is  equal  to  the  alternate  angle 
DEB,  and  the  angle  BGG  equal  to  BDE;  the  triangle  BGG  is 
equiangular  to  BDE:  therefore  as  GG  to  GB,  so  is  BD  to  DE; 
and  consequently  the  rectangle  GG,  DE  is  equal  to  the  rectangle 
GB,  BD  which  is  given,  because  its  sides  GB,  BD  are  given: 
therefore  the  rectangle  contained  by  the  excess  of  BA,  AG  and 
ihe  straight  line  DE  is  given. 
3  A 


370  Euclid's  data.  £ 

PROP.  XCIX.  95. 

If  from  a  given  point  in  the  diameter  of  a  circle 
given  in  position,  or  in  the  diameter  produced,  a 
straight  hne  be  drawn  to  any  point  in  the  circumfe- 
rence, and  from  that  point  a  straight  hne  be  drawn  at 
right  angles  to  the  first,  and  from  the  point  in  which 
this  meets  the  circumference  again,  a  straight  line  be 
drawn  parallel  to  the  first;  the  point  in  which  this  pa- 
rallel meets  the  diameter  is  given ;  and  the  rectangle 
contained  by  the  two  parallels  is  given. 

In  BC  the  diameter  of  the  circle  ABC  given  in  position,  or  in 
BC  produced,  let  the  given  point  D  be  taken,  and  from  D  let  a 
straight  line  DA  be  drawn  to  any  point  A  in  the  circumference, 
and  let  AE  be  drawn  at  right  angles  to  DA,  and  from  the  point  E 
where  it  meets  the  circumference  again  let  EF  be  drawn  parallel 
to  DA  meeting  BC  in  F;  the  point  F  is  given,  as  also  the  rectan- 
gle AD,  EF.  ' 

Produce  EF  to  the  circumference  in  G,  and  join  AG:  because 
GEA  is  a  right  angle,  the  straight  line  AG  is  (Cor.  5.  4.)  the  dia- 
meter of  the  circle  ABC;  and  BC  is  also  a  diameter  of  it;  there- 
fore the  point  H  where  they  meet  is  the  centre  of  the  circle,  and 
consequently  H  is  given:  and  the  point  D  is  given,  wherefore  DH 
is  given  in  magnitude:  and  because  AD  is  parallel  to  FG,  and 
GH  equal  to  HA;  DH  is  equal  (4»  6.)  to  HF,  and  AD  equal  to 
GF:  and  DH  is  given,  therefore  HF  is  given  in  magnitude;  and 

A  A 


it  is  also  given  in  position,  and  the  point  H  is  given,  therefore 
(30.  dat.)  the  point  F  is  given. 

And  because  the  straight  line  EFG  is  drawn  from  a  given  point 
F  without  or  within  the  circle  ABC  given  in  position,  therefore 
(95.  or  96.  dat.)  the  rectangle  EF,  FG  is  given:  and  GF  is  equal 
to  AD,  wherefore  the  rectangle  AD,  EF  is  given. 

PROP.  C.  Q. 

If  from  a  given  point  in  a  straight  line  given  in  po- 
sition, a  straight  hne  be  drawn  to  any  point  in  the  cir- 
cumference of  a  circle  given  in  position;  and  from  this 
point  a  straiglit  line  be  drawn,  making  with  the  first  an 


EUCLID  S    DATA. 


371 


angle  equal  to  the  difference  of  a  right  angle  and  the 
angle  contained  by  the  straight  line  given  in  position, 
and  the  straight  line  which  joins  the  given  point  and 
the  centre  of  the  circle;  and  from  the  point  in  which 
the  second  line  meets  the  circumference  again,  a  third 
straight  line  be  drawn,  making  with  the  second,  an  an- 
gle equal  to  that  which  the  first  makes  with  the  second: 
the  point  in  which  this  third  line  meets  the  straight  line 
given  in  position  is  given;  as  also  the  rectangle  con- 
tained by  the  first  straight  line  and  the  segment  of  the 
third  betwixt  the  circumference  and  the  straight  line 
given  in  position,  is  given. 

Let  the  straight  line  CD  be  drawn  from  the  given  point  C  in 
the  straight  line  AB  given  in  position,  to  the  circumference  of  the 
circle  DEF  given  in  position,  of  which  G  is  the  centre;  join  CG, 
and  from  the  point  D  let  DF  be  drawn,  making  the  angle  CDF 
equal  to  the  difference  of  a  right  angle  and  the  angle  BCG,  and 
from  the  point  F  let  FE  be  drawn,  making  the  angle  DFE  equal 
to  CDF,  meeting  AB  in  H:  the  point  H  is  given;  as  also  the 
rectangle  CD,  FH. 

Let  CD,  FH  meet  one  anotlier  in  the 
point  K,  from  which  draw  KL  perpendi- 
cular to  DF;  and  let  DC  meet  the  circum- 
ference again  in  M,  and  let  FH  meet  the 
same  in  E,  and  join  MG,  GF,  GH. 

Because  the  angles   MDF,  DFE  are  A 
equal  to  one  another,  the  circumferences 
MF,  DE  are  equal  (26.  3.);  and  adding 
or  taking  away  the  common  part  ME,  the 
circumference  DM  is  equal  to  EF;  there- 
fore the  straight  line  DM  is  equal  to  the 
straight  line  EF,  and  the  angle  GMD  to 
the  angle   (8.  1.)  GEE;   and    the  angles 
GMC,  GFH,are  equal  to  one  another,  be- 
cause they  are  either  the  same  with  the 
angles  GMD,  GEE  or  adjacent  to  them: 
and  because  the  angles  KDL,  LKD  are  to- 
gether equal  (32.  1.)  to  a  right  angle,  that 
is,  by  the  hypothesis,  to  the  angles  KDL, 
GCB;  the  angle  GCB,  or  GCH,  is  equal 
to  the  angle  (LKD,  that  is  to  the  angle) 
LKF  or  GHK:  therefore  the  points  C,  K, 
H,  G  are  in  the  circumference  of  a  circle; 
and  the  angle  GCK  is  therefore  equal  to     A  C  H      B 

the  angle  GHF;  and  the  angle  GMC  is  equal  to  GFH,  and  the 
straight  line  GM  to  GF;  therefore  (26.  1.)  CG  is  equal  to  GH,  and 
CM  to  HE:  and  because  CG  is  equal  to  GH,  the  angle  GCH  is 
equal  to  GHC;  but  the  angle   GCH  is   given:   therefore  GHC  is 


K       D 


•H 


372  Euclid's  data.  N 

given,  and  consequently  the  angle  CGH  is  given^  and  CG  is  given 
in  position,  and  the  point  G^  therefore  (32.  dat.)  GH  is  given  in 
position;  and  CB  is  also  given  in  position,  wherefore  the  point  H 
is  given. 

And  because  HF  is  equal  to  CM,  the  rectangle  DC,  FH  is  equal 
to  DC,  CM:  but  DC,  CM  is  given  (95.  or  96.  dat.),  because  the 
point  C  is  given,  therefore  the  rectangle  DC,  FH  is  given. 


FINIS. 


NOTES  ON  EUCLID'S  DATA. 


DEFINITION  II. 

This  is  made  more  explicit  than  in  the  Greek  text,  to  prevent  a 
mistake  which  the  author  of  the  second  demonstration  of  the  24th 
proposition  in  the  Greek  edition  has  fallen  into,' of  thinking  that  a 
ratio  is  given  to  which,  another  ratio  is  shown  to  be  equal,  though 
this  other  be  not  exhibited  in  given  magnitudes.  See  the  Notes 
on  that  proposition,  which  is  the  13th  in  this  edition.  Besides, 
by  this  definition,  as  it  is  now  given,  some  propositions  are  demon- 
strated, which,  in  the  Greek,  are  not  so  well  done  by  help  of  prop.  2. 

D^F.  IV. 

In  the  Greek  text,  def.  4,  is  thus;  "  Points,  lines,  spaces,  and  an- 
gles are  said  to  be  given  in  position  which  have  always  the  same 
situation;"  but  this  is  imperfect  and  useless,  because  there  are  in- 
numerable cases  in  which  things  may  be  given  according  to  this 
definition,  and  yet  their  position  cannot  be  found;  for  instance,  let 
the  triangle  ABC  be  given  in  position,  and  let  it  be  proposed  to 
draw  a  straight  line  BD  from  the  angle  at  B  to  the  opposite  side 
AC,  which  shall  cut  off  the  angle  DBC,  A 

which  shall  be  the  seventh  part  of  the  an-  -^ 

gle  ABC;  suppose  this  is  done,  therefore  ^y^    \ 

the  straight  line  BD  is  invariable  in  its  po-  ^     \     D 

sition,  that  is,  has  always  the  same  situation;    B     -^^^rzi^iil-L-L C 

for  any  other  straight  line  drawn  from  the  point  B  on  either  side 
of  BD  cuts  off  an  angle  greater  or  lesser  than  the  seventh  part  of 
the  angle  ABC;  therefore,  according  to  this  definition,  the  straight 
line  BD  is  given  in  position,  as  also  (28.dat.)  the  point  D  in  which  it 
meets  the  straight  line  AC  which  is  given  in  position.  But  from 
the  things  here  given,  neither  the  straight  line  BD  nor  the  point  D 
can  be  found  by  the  help  of  Euclid's  Elements  only,  by  which  every 
thing  in  his  Data  is  supposed  may  be  found.  This  definition  is 
therefore  of  no  use.  We  have  amended  it  by  adding,  '^  and  which 
are  either  actually  exhibited  or  can  be  found;"  for  nothing  is  to  be 
reckoned  given  which  cannot  be  found,  or  is  not  actually  exhibited. 

The  definition  of  an  angle  given  by  position  is  taken  out  of  the 
4th,  and  given  more  distinctly  by  itself  in  the  definition  marked  A. 

DEF.  XI.  XII.  XIII.  XIV.  XV. 

The  11th  and  12th  are  omitted,  because  they  cannot  be  given  in 
English  so  as  to  have  any  tolerable  sense:  and,  therefore,  where- 
ever  the  terms  defined  occur,  the  words  which  ex|5ress  their  mean- 
ing are  made  use  of  in  their  place. 


374  NOTES  ON  euolid's  d'ata. 

The  13th,  14th,  and  15th  are  omitted,  as  being  of  no  use. 

It  is  to  be  observed  in  general  of  the  Data  in  this  book,  that 
they  are  to  be  understood  to  be  given  geometrically,  not  always 
arithmetically;  that  is,  they  cannot  always  be  exhibited  in  num- 
bers; for  instance,  if  the  side  of  a  square  be  given,  the  ratio  of  it 
to  its  diameter  is  given  (44.  dat.)  geometrically,  but  not  in  num- 
bers; and  the  diameter  is  given  (2.  dat.);  but  though  the  number 
of  any  equal  parts  in  the  side  be  given,  for  example,  10,  the  num- 
ber of  them  in  the  diameter  cannot  be  given:  and  the  like  holds 
in  many  other  cases. 

PROPOSITION.  I. 

In  this  it  is  shown,  that  A  is  to  B,  as  C  to  D;  from  this,  that 
A  is  to  C,  as  B  to  D;  and  then  by  permutation:  but  it  follows  di- 
rectly, without  these  two  steps,  from  7^5. 

PROP.  II. 

The  limitation  added  at  the  end  of  this  proposition  between  the 
inverted  commas  is  quite  necessary,  because  without  it  the  propo- 
sition cannot  always  be  demonstrated:  for  the  author  having  said,* 
"because  A  is  given,  a  magnitude  equal  to  it  can  be  found  (1.  def.); 
let  this  be  C;  and  because  the  ratio  of  A  to  B  is  given,  a  ratio 
which  is  the  same  to  it  can  be  found  (2.  def.),"  adds,  "Let  it  be 
found,  and  let  it  be  the  ratio  of  C  to  A."  Now,  from  the  second 
definition  nothing  more  follows,  than  that  some  ratio,  suppose  the 
ratio  of  E  to  Z,  can  be  found,  which  is  the  same  with  the  ratio  of 
A  to  B;  and  when  the  author  supposes  that  the  ratio  of  C  to  A, 
which  is  also  the  same  with  the  ratio  of  A  to  B  can  be  found,  he 
necessarily  supposes  that  to  the  three  magnitudes  E,  Z,  C,  a  fourtJa 
proportional  A  may  be  found;  but  this  cannot  always  be  done  by 
the  Elements  of  Euclid;  from  which  it  is  plain  Euclid  must  have 
understood  the  proposition  under  the  limitation  which  is  now 
added  to  his  text.     An  example  will  A  B        A 

make  this  clear:  let  A  be  a  given  angle, 
and  B  another  angle  to  which  A  has  a 
given  ratio:  for  instance,  the  ratio  of  the 
given  straight  line  E  to  the  given  one 
Z;  then,  having  found  an  angle  C  equal 
to  A,  how  can  the  angle  a  be  found  to 
which  C  has  the  same  ratio  that  E  has 
to  Z?  Certainly  no  way,  until  it  be 
shown  how  to  find  an  angle  to  which  a 
given  angle  has  a  given  ratio,  which 
cannot  be  done  by  Euclid's  Elements,  nor  probably  by  any  Geome- 
try known  in  his  time.  Therefore,  in  all  the  propositions  of  this 
book  which  depend  upon  this  second,  the  above  mentioned  limita- 
tion must  be  understood,  though  it  be  not  explicitly  mentioned. 

PROP.  V. 

The  order  of  the  propositions  in  the  Greek  text,  between  prop. 
4,  and  prop.  2.5.  is  now  changed  into  another  which  is  more  natu- 

^  See  Dr.  Gregory's  edition  of  the  Data. 


NOTES  ON  EUCLID  S  DATA.  375 

ral,  by  placing  tliose  which  are  more  simple  before  those  which 
are  more  complex;  and  by  placing  together  those  which  are  of 
the  same  kind,  some  of  which  were  mixed  among  others  of  a  dif- 
ferent kind.  Thus,  prop.  12,  in  the  Greek,  is  now  made  the  5th, 
and  those  which  were  the  32d  and  23d  are  made  the  11th  and 
12th,  as  they  are  more  simple  than  the  propositions  concerning 
magnitudes,  the  excess  of  one  of  which  above  a  given  magnitude 
has  a  given  ratio  to  the  other,  after  which  these  two  M'ere  placed^ 
and  the  24th  in  the  Greek  text  is,  for  the  same  reason,  made  the 
13th. 

PROP.  VI.  VII. 

These  are  universally  true,  though  in  the  Greek  text  they  are 
demonstrated  by  prop.  2,  which  has  a  limitation^  they  are  there- 
fore now  shown  without  it. 

PROP.  XII. 

In  the  23d  prop,  in  the  Greek  text,  which  here  is  the  12th,  the 
words,  "  y^i}  rH<i  civrs(i  JV,"  are  wrong  translated  by  Claud.  Hardy, 
in  his  edition  of  Euclid's  Data,  printed  at  Paris,  anno  1625,  which 
was  the  first  edition  of  the  Greek  text;  and  Dr.  Gregory  follows 
him  in  translating  them  by  the  words,  "  etsi,  non  easdera,"  as  if 
the  Greek  had  been  si  kxi  f^tjla^  uvm^;^  as  in  prop.  9.  of  the  Greek 
text.  Euclid's  meaning  is,  that  the  ratios  mentioned  in  the  pro- 
position must  not  be  the  same;  for,  if  they  were,  the  proposition 
would  not  be  true.  Whatever  ratio  the  whole  has  to  the  whole, 
if  the  ratios  of  the  parts  of  the  first  to  the  parts  of  the  other  be  the 
same  with  this  ratio,  one  part  of  the  first  may  be  double,  triple.  See. 
of  the  other  part  of  it,  or  have  any  other  ratio  to  it,  and  conse- 
quently cannot  have  a  given  ratio  to  it;  wherefore,  these  words 
must  be  rendered  by  "  non  autem  easdem,"  but  not  the  same 
ratios,  as  Zambertus  has  translated  them  in  his  edition. 

PROP.  XIII. 

Some  very  ignorant  editor  has  given  a  second  demonstration  of 
this  proposition  in  the  Greek  text,  which  has  been  as  ignorantly 
kept  in  by  Claud.  Hardy  and  Dr.  Gregory,  and  has  been  retained 
in  the  translations  of  Zambertus  and  others;  Carolus  Renaldinus 
gives  it  only:  the  author  of  it  has  thought  that  a  ratio  was  given, 
if  another  ratio  could  be  shown  to  be  the  same  to  it,  though  this 
last  ratio  be  not  found:  but  this  is  altogether  absurd,  because 
from  it  would  be  deduced,  that  the  ratio  of  the  sides  of  any  two 
squares  is  given,  and  the  ratio  of  the  diameters  of  any  tv>'o  cir- 
cles, &c.  And  it  is  to  be  observed,  that  the  moderns  frequently 
take  given  ratios,  and  ratios  that  are  always  the  same,  for  one 
and  the  same  thing;  and  Sir  Isaac  Newton  has  fallen  into  this 
mistake  in  the  17th  lemma  of  his  Principia,  edit.  1713,  and  in 
other  places:  but  this  should  be  carefully  avoided,  as  it  may  lead 
into  other  errors. 

PROP.  XIV.   XV. 

Euclid,  in  this  book,  has  several  propositions  concerning  mag- 
nitudes, the  excess  of  one  of, which  above  a  given  magnitude  has 


376  NOTES  ON  Euclid's  data. 

a  given  ratio  to  the  other;  but  he  has  given  none  concei'ning 
magnitudes  whereof  one  together  with  a  given  magnitude  has  a 
given  ratio  to  the  other;  though  these  last  occur  as  frequently  in 
the  solution  of  problems  as  the  first;  the  reason  of  which  is,  that 
the  last  may  be  all  demonstrated  by  help  of  the  first;  for,  if  a 
magnitude,  together  v/ith  a  given  magnitude,  has  a  given  ratio  to 
another  magnitude,  the  excess  of  this  other  above  a  given  magni- 
tude shall  have  a  given  ratio  to  the  first,  and  on  the  contrary;  as 
we  have  demonstrated  in  prop.  !4.     And  for  a  like  reason  prop. 
15   has  been   added,  to  the  Data.     One  example  will  make  the 
thing  clear:  suppose  it  were  to  be  demonstrated,  that  if  a  magni- 
tude A  together  with   a  given  magnitude  has  a  given  ratio  to 
another  magnitude  B,  that  the  two  magnitudes  A  and  B,  together  ?: 
with  a  given  magnitude,  have  a  given  ratio  to  that  other  inagni- 
tude  B;  which  is  the  same  proposition  with  respect  to  the  last 
kind  of  magnitudes  above  mentioned,  that  the  first  part  of  prop. 
16,  in  this  edition,  is  in  respect  of  the  first  kind:  this  is  shown 
thus;  from  the  hypothesis,  and  by  the  first  part  of  prop.  14,  the 
excess  of  B  above  a  given  magnitude  has  unto  A  a  given  ratio; 
and,  therefore,  by  the  first  part  of  prop.  17,  the  excess  of  B  above 
a  given  magnitude  has  unto  B  and  A  together  a  given  ratio;  and 
by  the  second  part  of  prop.  14,  A  and  B  together  with  a  given 
magnitude  has  unto  B  a  given  ratio;  which  is  the  thing  that  was 
to  be  demonstrated.     In  like  manner  the  other  propositions  con- 
cerning the  last  kind  of  magnitudes  may  be  shown. 

PROP.  XVI.   XVII. 

In  the  third  part  of  prop.  10,  in  the  Greek  text,  which  is  the 
16th  in  this  edition,  after  the  ratio  of  EC  to  CB  has  been  shown 
to  be  given;  from  this,  by  inversion  and  conversion,  the  ratio  of 
'  BC  to  BE  is  demonstrated  to  be  given;  but  without  these  two 
steps,  the  conclusion  should  have  been  made  only  by  citing  the  6th 
proposition.  And  in  like  manner,  in  the  first  part  of  prop.  11,  in 
the  Greek,  which  in  this  edition  is  the  17th,  from  the  ratio  of  DB 
to  BC  being  given,  the  ratio  of  DC  to  DB  is  shown  to  be  given  by 
inversion  and  composition,  instead  of  citing  prop.  7,  and  the  same 
fault  occurs  in  the  second  part  of  the  same  prop.  11. 

PROF.  XXI.    XXII. 

These  now  are  added,  as  being  wanting  to  complete  the  subject 
treated  of  in  the  four  preceding  propositions. 

PROP.  XXIII. 

This,  which  is  prop.  20,  in  the  Greek  text,  was  separated  from 
prop.  14,  15,  16,  in  that  text,  after  which  it  should  have  been  im- 
mediately placed,  as  being  of  the  same  kind:  it  is  now  put  into  its 
proper  place;  but  prop.  21,  in  the  Greek,  is  left  out,  as  being  the 
same  with  prop.  14,  in  that  text,  which  is  here  prop.  18. 

PROP.  XXIV. 

This,  which  is  prop.  13,  in  the  Greek,  is'now  put  into  its  proper 


NOTES  ON  Euclid's  data.  377 

place,  having  been  disjoined  from  the  three  following  it  in  this 
edition,  which  are  of  the  same  kind. 

PROP.  XXVIII. 

This,  which  in  the  Greek  text  is  prop.  25,  and  several  of  the 
following  propositions,  are  there  deduced  from  def.  4,  which  is 
not  sufficient,  as  has  been  mentioned  in  the  note  on  that  definition: 
they  are  therefore  now  shown  more  explicitly. 

PROP.  XXXIV.  XXXVI. 

Each  of  these  has  a  determination,  which  is  now  added,  which 
occasions  a  change  in  their  demonstrations. 

PROP.  XXXVII.  XXXIX,  XL.  XLI. 

The  35th  and  36th  propositions  in  the  Greek  text  are  joined 
into  one,  which  makes  the  39th  in  this  edition,  because  the  same 
enunciation  and  demonstration  serves  both:  and  for  the  same 
reason  prop.  37,  38,  in  the  Greek,  are  joined  into  one,  which  here 
is  the  40th. 

Prop.  37  is  added  to  the  Data,  as  it  frequently  occurs  in  the 
solution  of  problems;  and  prop.  41  is  added  to  complete  the  rest. 

PROP.  XLII. 

This  is  prop.  39,  in  the  Greek  text,  where  the  whole  construc- 
tion of  prop.  22,  of  book  1,  of  the  Elements,  is  put,  without  need, 
into  the  demonstration,  but  is  now  only  cited. 

PROP.  XLV. 

This  is  prop.  42,  in  the  Greek,  where  the  three  straight  lines 
made  use  of  in  the  construction  are  said,  but  not  shown,  to  be 
such  that  any  two  of  them  is  greater  than  the  third,  which  is 
now  done. 

PROP.  XLVII. 

'  This  is  prop.  44,  in  the  Greek  text;  but  the  demonstration  of 
it  is  changed  into  another,  wherein  the  several  cases  of  it  are 
shown,  which,  though  necessary,  is  not  done  in  the  Greek. 

PROP.  XLVIII. 

There  are  two  cases  in  this  proposition,  arising  from  the  two 
cases  of  the  third  part  of  prop.  47,  on  which  the  48th  depends; 
and  in  the  composition  these  two  cases  are  explicitly  given. 

PROP.  LII. 

The  construction  and  demonstration  of  this,  which  is  prop.  48, 
in  the  Greek,  are  made  something  shorter  than  in  that  text. 

PROP.  LIII. 

Prop.  63,  in  the  Greek  text,  is  omitted,  being  only  a  case  of 
prop.  49,  in  that  text,  which  is  prop.  53,  in  this  edition. 
3  B 


378  NOTES  ON  Euclid's  data. 

PROP.  LVIII. 

This  is  not  in  the  Greek  text,  but  its  demonstration  is  contain- 
ed in  that  of  the  first  part  of  prop.  54,  in  that  text;  which  propo- 
sition is  concerning  figures  that  are  given  in  species:  this  58th 
is  true  of  similar  figures,  though  they  be  not  given  in  species^ 
and  as  it  frequently  occurs,  it  was  necessary  to  add  it. 

PROP.  LIX.  LXI. 

This  is  the  54th  in  the  Greek;  and  the  77th  in  the  Greek,  be- 
ing the  very  same  with  it,  is  left  out,  and  a  shorter  demonstration 
is  given  of  prop.  61. 

PROP.  LXII. 

This,  which  is  most  frequently  useful,  is  not  in  the  Greek,  and 
is  necessary  to  prop.  87,  88,  in  this  edition,  as  also,  though  not 
mentioned,  to  prop.  86,  87,  in  the  former  editions.  Prop.  66,  in 
the  Greek  text,  is  made  a  corollary  to  it. 

PROP.  LXIV. 

This  contains  both  prop.  74,  and  73,  in  the  Greek  text;  the 
first  case  of  the  74th  is  a  repetition  of  prop.  56,  from  which  it  is 
separated  in  that  text  by  many  propositions;  and  as  there  is  no 
order  in  these  propositions,  as  they  stand  in  the  Greek,  they  are 
now  put  into  the  order  which  seemed  most  convenient  and  na- 
tural. 

The  demonstration  of  the  first  part  of  prop.  12>^  in  the  Greek, 
is  grossly  vitiated.  Dr.  Gregory  says,  that  the  sentences  he  has 
enclosed  betwixt  two  stars  are  superfluous,  and  ought  to  be  can- 
celled; but  he  has  not  observed,  that  what  follows  them  is  absurd, 
being  to  prove  that  the  ratio  [see  his  figure]  of  Ar  to  rK  is  given, 
which  by  the  hypothesis  at  the  beginning  of  the  proposition  is 
expressly  given:  so  that  the  whole  of  this  part  was  to  be  altered, 
which  is  done  in  this  prop.  64. 

PROP.  LXVII.  LXVIII. 

Prop.  70,  in  the  Greek  text,  is  divided  into  these  two,  for  the 
sake  of  distinctness;  and  the  demonstration  of  the  67th  is  render- 
ed shorter  than  that  of  the  first  part  of  prop.  70,  in  the  Greek, 
by  means  of  prop.  23,  of  book  6,  of  the  Elements. 

PROP.  LXX. 

This  is  prop.  62,  in  the  Greek  text;  prop.  78,  in  that  text,  is 
only  a  particular  case  of  it,  and  is  therefore  omitted. 

Dr.  Gregory,  in  the  demonstration  of  prop.  62,  cites  the  49th 
prop.  dat.  to  prove  that  the  ratio  of  the  figure  AEB  to  the  pa- 
rallelogram AH  is  given;  whereas  this  v/as  shov/n  a  few  lines 
before:  and  besides,  the  49th  prop,  is  not  applicable  to  these  two 
figures;  because  AH  is  not  given  in  species,  but  is,  by  the  step 
for  which  the  citation  is  brought,  proved  to  be  given  in  species. 

PROP.  Lxxni. 

Prop.  83,  in  the  Greek  text,  is  neither  well  enunciated  nor 
demonstrated.    The  73d,  which  in  this  edition  is  put  in  place  of 


«*;;,' 


NOTES  ON  Euclid's  data.  379 


it,  is  really  the  same,  as  will  appear  by  considering  [see  Dr« 
Gregory's  edition]  that  A,  B,  r,  E  in  the  Greek  text  are  four 
proportionals;  and  that  the  proposition  is  to  show  that  A,  which 
has  a  given  ratio  to  E,  is  to  r,  as  B  is  to  a  straight  line  to  which 
A  has  a  given  ratio;  or,  by  inversion,  that  r  is  to  a,  as  a  straight 
line  to  which  A  has  a  given  ratio  is  to  B;  that  is,  if  the  propor- 
tionals be  placed  in  this  order,  viz.  r,  E,  A,  B,  that  the  first  r  is 
to  A  to  which  the  second  E  has  a  given  ratio,  as  a  straight  line 
to  which  the  third  A  has  a  given  ratio  is  to  the  fourth  B;  which 
is  the  enunciation  of  this  73d,  and  was  thus  changed  that  it  might 
be  made  like  to  that  of  prop.  72,  in  this  edition,  which  is  the  82d 
in  the  Greek  text:  and  the  demonstration  of  prop.  73  is  the  same 
with  that  of  prop.  72,  only  making  use  of  prop.  23,  instead  of 
prop.  22,  of  book  5,  of  the  Elements. 

PROP.  LXXVII. 

This  is  put  in  place  of  prop.  79,  in  the  Greek  text,  which  is 
not  a  datum,  but  a  theorem  premised  as  a  lemma  to  prop.  80  in 
that  text:  and  prop.  79  is  made  cor.  1  to  prop.  77 ^  in  this  edi- 
tion. CI.  Hardy,  in  his  edition  of  the  Data,  takes  notice,  that  in 
prop.  80,  of  the  Greek  text,  the  parallel  KL  in  the  figure  of  prop. 
77,  in  this  edition,  must  meet  the  circumference,  but  does  not 
demonstrate  it,  which  is  done  here  at  the  end  of  cor.  3,  prop.  77^ 
in  the  construction  for  finding  a  triangle  similar  to  ABC. 

PROP.  LXXVIII. 

The  demonstration  of  this,  which  is  prop.  80,  in  the  Greek,  is 
rendered  a  good  deal  shorter  by  help  of  prop.  77. 

PROP.  LXXIX.  LXXX.  LXXXI. 

These  are  added  to  Euclid's  Data,  as  propositions  which  are 
often  useful  in  the  solution  of  problems. 

PROP.  LXXXIT. 

This,  which  is  prop.  60,  in  the  Greek  text,  is  placed  before 
the  83d  and  84th,  which,  in  the  Greek,  are  the  58th  and  59th,  be- 
cause the  demonstration  of  these  two  in  this  edition  are  deduced 
from  that  of  prop.  82,  from  which  they  naturally  follow. 

PROP.  LXXXVIIl.  XC. 

Dr.  Gregory,  in  his  preface  to  Euclid's  Works,  which  he  pub- 
lished at  Oxford  in  1703,  after  having  told  that  he  had  supplied 
the  defects  of  the  Greek  text  of  the  Data  in  innumerable  places 
from  several  manuscripts,  and  corrected  CI.  Hardy's  translation 
by  Mr.  Bernard's;  adds,  that  the  86th  theorem,  "  or  proposition," 
seemed  to  be  remarkably  vitiated,  but  which  could  not  be  re- 
stored by  help  of  the  manuscripts;  then  he  gives  three  different 
translations  of  it  in  Latin,  according  to  which,  he  thinks  it  may 
be  read;  the  two  first  have  no  distinct  meaning,  and  the  third, 
which  he  says  is  the  best,  though  it  contains  a  true  proposition, 
which  is  the  9th  in  this  edition,  has  no  connexion  in  the  least 
with  the  Greek  text.  And  it  is  strange  that  Dr.  Gregory  did  not 
observe,  that,  if  prop.  86  was  changed  into  this,  the  demonstra- 


380  NOTES  ON  EUCLID*S    DA^ A. 

tion  of  the  86th  must  be  cancelled,  and  another  put  in  its  place: 
but  the  truth  is,  both  the  enunciation  and  the  demonstration  of 
prop.  86  are  quite  entire  and  right,  only  prop.  87,  which  is  more 
simple,  ought  to  have  been  placed  before  it;  and  the  deficiency 
"which  the  doctor  justly  observes  to  be  in  this  part  of  Euclid's 
Data,  and  which,  no  doubt,  is  owing  to  the  carelessness  and  ig- 
norance of  the  Greek  editors,  should  have  been  supplied,  not  by 
changing  prop.  86,  which  is  both  entire  and  necessary,  but  by 
adding  the  two  propositions,  which  are  the  88th  and  90th  in  this 
edition. 

PROP.  XCVIII.  C. 

These  were  communicated  to  me  by  two  excellent  geometers, 
the  first  of  them  by  the  Right  Honourable  the  Earl  of  Stanhope, 
and  the  other  by  Dr.  Matthew  Stewart;  to  which  I  have  added 
the  demonstrations. 

Though  the  order  of  the  propositions  has  been  in  many  places 
changed  from  that  in  former  editions,  yet  this  will  be  of  little 
disadvantage,  as  the  ancient  geometers  never  cite  the  Data,  and 
the  moderns  very  rarely. 


AS  that  part  of  the  composition  of  a  problem  which  is  its  con- 
struction may  not  be  so  readily  deduced  from  the  analysis  by  be- 
ginners; for  their  sake  the  following  example  is  given,  in  which 
the  deviation  of  the  several  parts  of  the  construction  from  the 
analysis  is  particularly  shown,  that  they  may  be  assisted  to  do  the 
like  in  other  problems. 

PROBLEM. 

Having  given  the  magnitude  of  a  parallelogram,  the  angle  of 
which  ABC  is  given,  and  also  the  excess  of  the  square  of  its  side 
BC  above  the  square  of  the  side  AB;  to  find  its  sides,  and  de- 
scribe it. 

The  analysis  of  this  is  the  same  with  the  demonstration  of  the 
87th  prop,  of  the  Data,  and  the  construction  that  is  given  of  the 
problem  at  the  end  of  that  proposition  is  thus  derived  from  the 
analysis. 

LetEFG  be  equal  to  the  given  angle  ABC,  and  because  in  the 
analysis  it  is  said  that  the  ratio  of  the  rectangle  AB,  BC  to  the 
parallelogram  AC  is  given  by  the  62d  prop,  dat.,  therefore,  from 
a  point  inFE,the  perpendicular  EG  is  drawn  to  FG,  as  the  ratio 
of  FE  to  EG  is  the  ratio  of  the  rectangle  AB,  BC  to  the  parallelo- 

M 


B      PD     C  F  G       L     O         HN 

gram  AC,  by  what  is  shown  at  the  end  of  prop.  62.     Next,  the 


NOTES  ON  EUCLID  S  DATA.  381 

magnitude  of  AC  is  exhibited  by  making  the  rectangle  EG,  GH 
equal  to  h;  and  the  given  excess  of  the  square  of  BC  above  the 
square  of  BA,  to  which  excess  the  rectangle  CB,  BL  is  equal,  is 
exhibited  by  the  rectangle  HG,  GL:  then,  in  the  analysis,  the 
rectangle  AB,  BC  is  said  to  be  given,  and  this  is  equal  to  the 
rectangle  FE,  GH,  because  the  rectangle  AB,  BC  is  to  the  pa- 
rallelogram AC,  as  (FE  to  EG,  that  is,  as  the  rectangle)  FE, 
GH  to  EG,  GH;  and  the  parallelogram  AC  is  equal  to  the  rect- 
angle EG,  GH,  therefore  the  rectangle  AB,  BC,  is  equal  to  FE, 
GH:  and  consequently  the  ratio  of  the  rectangle  CB,  BD,  that 
is,  of  the  rectangle  HG,  GL,  to  AB,  BC,  that  is  of  the  straight 
line  DB  to  BA,  is  the  same  with  the  ratio  (of  the  rectangle  GL, 
GH  to  FE,  GH,  that  is)  of  the  straight  line  GL  to  FE,  which 
ratio  of  DB  to  BA  is  the  next  thing  said  to  be  given  in  the  ana- 
lysis: from  this  it  is  plain  that  the  square  of  FE  is  to  the  square 
of  GL,  as  the  square  of  BA,  which  is  equal  to  the  rectangle  BC, 
CD,  is  to  the  square  of  BD:  the  ratio  of  which  spaces  is  the 
next  thing  said  to  be  given:  and  from  this  it  follows  that  four 
times  the  square  of  FE  is  to  the  square  of  GL,  as  four  times  the 
rectangle  BC,  CD  is  to  the  square  of  BD;  and,  by  composition, 
four  times  the  square  of  FE  together  with  the  square  of  GL,  is 
to  the  square  of  GL,  as  four  times  the  rectangle  BC,  CD,  toge- 
ther with  the  square  of  BD,  is  to  the  square  of  BD,  that  is  (8.  6.) 
as  the  square  of  the  straight  lines  BC,  CD  taken  together  is  to 
the  square  of  BD,  which  ratio  is  the  next  thing  said  to  be  given 
in  the  analysis:  and  because  four  times  the  square  of  FE  and  the 
square  of  GL  are  to  be  added  together;  therefore  in  the  perpen- 
dicular EG  there  is  taken  KG  equal  to  FE,  and  MG  equal  to  the 
double  of  it,  because  thereby  the  squares  of  MG,  GL,  that  is, 
joining  ML,  the  square  of  ML  is  equal  to  four  times  the  square 
of  FE  and  to  the  square  of  GL:  and  because  the  square  of  ML 
is  to  the  square  of  GL,  as  the  square  of  the  straight  line  made  up 
of  BC  and  CD  is  to  the  square  of  BD,  therefore  (22.  6.)  ML  is 
to  LG,  as  BC  together  with  CD  is  to  BD;  and,  by  composition, 
ML  and  LG  together,  that  is,  producing  GL  to  N,  so  that  ML 
be  equal  to  LN,  the  straight  line  NG  is  to  GL,  as  twice  BC  is  to 
BD;  and  by  taking  GO  equal  to  the  half  of  NG,  GO  is  to  GL,  as 
BC  to  BD,  the  ratio  of  which  is  said  to  be  given  in  the  analysis: 
and  from  this  it  follows,  that  the  rectangle  HG,  GO  is  to  HG, 
GL,  as  the  square  of  BC  to  the  rectangle  CB,  BD,  v/hich  is  equal 
to  the  rectangle  HG,  GL;  and  therefore  the  square  of  BC  is  equal 
to  the  rectangle  HG,  GO;  and  BC  is  consequently  found  by 
taking  a  mean  proportional  betwixt  HG  and  GO,  as  is  said  in 
the  construction:  and  because  it  was  shown  that  GO  is  to  GL,  as 
BC  to  BD,  and  that  now  the  three  first  are  found,  the  fourth  BD 
is  found  by  12.  6.  It  was  likewise  shown  that  LG  is  to  FG,  or 
GK,  as  DB  to  BA,  and  the  three  first  are  nov/  found,  and  there- 
by the  fourth  BA.  Make  the  angle  ABC  equal  to  EFG,  and 
complete  the  parallelogram  of  which  the  sides  are  AB,  GC,  and 
the  construction  is  finished;  the  rest  of  the  composition  contains 
the  demonstration.  * 


382  NOTES  ON  Euclid's  data'. 

AS  the  propositions  from  the  13th  to  the  28th  may  be  thought 
by  beginners  to  be  less  useful  than  the  rest,  because  they  cannot 
so  readily  see  how  they  are  to  be  made  use  of  in  the  solution  of 
problems;  on  this  account  the  two  following  problems  are  added, 
to  show  that  they  are  equally  useful  with  the  other  propositions, 
and  from  which  it  may  be  easily  judged  that  many  other  problems 
depend  upon  these  propotitions. 

PROBLEM  I. 

To  find  three  straight  fines  such,  that  the  ratio  of 
the  first  to  the  second  is  given;  and  if  a  given  straight 
fine  be  taken  from  the  second,  the  ratio  of  the  remain- 
der to  the  third  is  given;  also  the  rectangle  contained 
by  the  first  and  third  is  given. 

Let  AB  be  the  first  straight  line,  CD  the  second,  and  EF  the 
third:  and  because  the  ratio  of  AB  to  CD  is  given,  and  that  if  a 
given  straight  line  be  taken  from  CD,  the  ratio  of  the  remainder 
to  EF  is  given:  therefore  (24.  dat.)  the  excess  of  the  first  AB 
above  a  given  straight  line  has  a  given  ratio  to  the  third  EF;  let 
BH  be  that  given  straight  line;  therefore  AH,      *  KB 

the  excess  of  AB  above  it,  has  a  given  ratio  to I 

EF;  and  consequently  (l.  6.)  the  rectangle  BA,  ' 

AH,  has  a  given  ratio  to  the  rectangle  AB,  EF,     C         G  D 

which  last  rectangle  is  given  by  the  hypothesis;     1 

therefore  (2.  dat.)   the    rectangle   BA,  AH   is     E     p 

given,  and  BH  the  excess  of  its  sides  is  given;     . 

wherefore  the  sides  AB,  AH  are  given  (85.  dat.):     -j^     m  m  L    O 

and  because  the  ratios  of  AB  to  CD,  and  of  AH     i , i 

to  EF  are  given,  CD  and  EF  are  (2.  dat.)  given.  '    '     ' 

The  Composition. 

Let  the  given  ratio  of  KL  to  KM  be  that  which  AB  is  required 
to  have  to  CD;  and  let  DG  be  the  given  straight  line  which  is  to 
be  taken  from  CD,  and  let  the  given  ratio  of  KM  to  KN  be  that 
which  the  remamder  must  have  to  EF;  also  let  the  given  rect- 
angle NK,  KO  be  that  to  which  the  rectangle  AB,  EF  is  required 
to  be  equal:  find  the  given  straight  line  BH  which  is  to  be  taken 
from  AB,  which  is  done,  as  plainly  appears  from  prop.  24.  dat.  by 
making  as  KM  to  KL,  so  GD  to  HB.  To  the  given  straight  line 
BH  apply  (29.  6.)  a  rectangle  equal  to  LK,  KO  exceeding  by  a 
square,  and  let  BA,  AH  be  its  sides:  then  is  AB  the  first  of  the 
straight  lines  required  to  be  found,  and  by  making  as  LK  to  KM, 
so  AB  to  DC,  DC  will  be  the  second:  and  lastly,  make  as  KM  to 
KN,  so  CG  to  EF,  and  EF  is  the  third. 

For  as  AB  to  CD,  so  is  HB  to  GD,  each  of  these  ratios  being 
the  same  with  the  ratio  of  LK  to  KM;  therefore  (19.  5.)  AH  is  to 
CG,  as  (AB  to  CD,  that  is,  as)  LK  to  KM;  and  as  CG  to  EF,  so 
is  KM  to  KN:  wherefore,  ex  aequali^  as  AH  to  EF,  so  is  LK  to 
KN:  and  as  the  rectangle  BA,  AH  to  the  rectangle  BA,EF,  so  is 
(1.6.)  the  rectangle  LK,  KO  to  the  rectangle  KN,  KO:  and  by  the 
construction,  the  rectangle  BA,  AH  is  equal  to  LK,  KO:  there- 


NOTES  ON  EUCLID  S  DATA. 


383 


fore  (14.  5.)  the  rectangle  AB,  EF  is  equal  to  the  given  rectangle 
NK,  KO:  and  AB  has  to  CD  the  given  ratio  of  KL  to  KMj  and 
from  CD  the  given  straight  line  GD  being  taken,  the  remain- 
der CG  has  to  EF  the  given  ratio  of  KM  to  KN.    Q.  E.  D. 

PROB.  II. 

To  find  three  straight  lines  such,  that  the  ratio  of  the 
first  to  the  second  is  given;  and  if  a  given  straight  hne 
be  taken  from  the  second,  the  ratio  of  the  remainder  to 
the  third  is  given;  also  tlie  sum  of  the  squares  of  the 
first  and  third  is  given. 

Let  AB  be  the  first  straight  line,  BC  the  second,  and  BD  the 
third:  and  because  the  ratio  of  AB  to  BC  is  given,  and  that  if  a 
given  straight  line  be  taken  from  BC,  the  ratio  of  the  remainder 
to  BD  is  given;  therefore  (24.  dat.)  the  excess  of  the  first  AB 
above  a  given  straight  line,  has  a  given  ratio  to  the  third  BD:  let 
AE  be  that  given  straight  line,  therefore  the  remainder  EB  has  a 
given  ratio  to  BD;  let  BD  be  placed  at  right  angles  to  EB,  and 
join  DE;  then  the  triangle  EBD  is  (44.  dat.)  given  in  species; 
wherefore  the  angle  BED  is  given:  let  AE,  which  is  given  in 
magnitude,  be  given  also  in  position,  as  also  the  point  E,  and  the 
straight  line  ED  will  be  given  (32.  dat.)  in  position:  join  AD,  and 
because  the  sum  of  the  squares  of  AB,  BD,  that  is  (47.  1.),  the 
square  of  AD  is  given,  therefore  the  straight  line  AD  is  given  in 
magnitude;  and  it  is  also  given  (34.  dat.)  in  position,  because 
from  the  given  point  A  it  is  drawn  to  the  straight  line  ED  given 
in  position:  therefore  the  point  D,  in  which  the  two  straight  lines 
AD,  ED  given  in  position  cut  one  another,  is  given  (28.  dat.):  and 
the  straight  line  DB  which  is  at  right  angles  to  AB  is  given  (33. 
dat.)  in  position,  and  AB  is  given  in  position,  therefore  (28.  dat.) 
the  point  B  is  given:  and  the  points  A,  D  are  given,  wherefore 
(29.  dat.)  the  straight  lines  AB,  BD  are  given;  and  the  ratio  of 
AB  to  BC  is  given,  and  therefore  (2.  dat.)  BC  is  given. 

The  Composition. 

Let  the  given  ratio  of  FG  to  GH  be  that  which  AB  is  required 
to  have  to  BC,  and  let  HK  be  the  given  straight  line  which  is  to 
be  taken  from  BC,  and  let  the  ratio  which  the  remainder  is  re- 


D 


B  N   M 


quired  to  have  to  BD,  be  the  given  ratio  of  HG  to  LG,  and  place 
GL  at  right  angles  to  FH,  and  join  LF,  LH:  next,  as  HG  is  to 
GF,  so  make  HK  to  AE;  produce  AE  to  N,  so  that  AN  be  the 


384  NOTES  ON  Euclid's  data. 

straight  line  to  the  square  of  which  the  sum  of  the  squares  of  AB^ 
BD  is  required  to  be  equal;  and  make  the  angle  NED  equal  to 
the  angle  GFL;  and  from  the  centre  A  at  the  distance  AN  de- 
scribe a  circle,  and  let  its  circumference  meet  ED  in  D,  and  draw 
DB  perpendicular  to  AN,  and  DM,  making  the  angle  BDM  equal 
to  the  angle  GLH.  Lastly,  produce  BM  to  C,  so  that  MC  be 
equal  to  HK;  then  is  AB  the  first,  BC  the  second,  and  BD  the 
third  of  the  straight  lines  that  were  to  be  found. 

For  the  triangles  EBD,  FGL,  as  also  DBM,  LGH  being  equi- 
angular, as  EB  to  BD,  so  is  FG  to  GL;  and  as  DB  to  BM,  so  is 
LG  to  GH;  therefore,  ex  eequali,  as  EB  to  BM,  so  is  (FG  to  GH, 
and  so  is)  AE  to  HK  or  MC;  wherefore  (12.  5.)  AB  is  to  BC,  as 
AE  to  HK,  that  is,  as  FG  to  GH,  that  is,  in  the  given  ratio;  and 
from  the  straight  line  BC  taking  MC,  which  is  equal  to  the  given 
straight  line  HK,  the  remainder  BM  has  to  BD  the  given  ratio 
of  HG  to  GL;  and  the  sum  of  the  squares  of  AB,  BD  is  equal 
(47.  1.)  to  the  square  of  AD  or  AN,  which  is  the  given  space. 
Q.  E.  D. 

I  believe  it  would  be  in  vain  to  try  to  deduce  the  preceding  con- 
struction from  an  algebraical  solution  of  the  problem. 


FINIS. 


■*';^  • 


THE 


ELEMENTS 


OF 


,i©^m 


^ 


TRIGON^OMETRY* 


3C 


^i§ 


« 


PLANE  TRIGONOMETRY. 


LEMMA  I.    FIG.  1. 

Let  ABC  be  a  rectilineal  angle;  if  about  the  point  B  as  a  cen- 
tre, and  wiih  any  distance  BA,  a  circle  be  descril)ed,  meeting  BA, 
BC,  the  straight  lines  including  the  angle  ABC  in  A,  C;  the 
angle  ABC  will  he  to  four  right  angles,  as  the  arch  AC  to  the 
whole  circurnfei-ence. 

Produce  AB  till  it  meet  the  circle  again  in  F,  and  through  B 
draw  DE  perpendicular  to  AB,  meeting  the  circle  in  D,  E. 

By  33.  6.  Elem.  the  angle  ABC  is  to  a  right  angle  ABD,  as 
the  arch  AC  to  the  arch  AD;  and  quadrupling  the  consequents, 
the  angle  ABC  will  be  to  four  right  angles,  as  the  arch  AC  to 
four  times  the  arch  AD,  or  to  the  whole  circumference. 

LEMMA  IL  FIG.  2. 

Let  ABC  be  a  plane  rectilineal  angle  as  before:  about  B  as  a 
centre,  with  any  two  distances  BD,  BA,  let  two  circles  be  de- 
scribed meeting  BA,  BC  in  D,  E,  A,  C;  the  arch  AC  will  be  to 
the  whole  circumference  of  which  it  is  an  arch,  as  the  arch  DE  is 
to  the  whole  circumference  of  which  it  is  an  arch. 

By  Lemma  1,  the  arch  AC  is  to  the  whole  circumference  of 
which  it  is  an  arch,  as  the  angle  ABC  is  to  four  right  angles; 
and  by  the  same  Lemma  1,  the  arch  DE  is  to  the  whole  circum- 
ference of  which  it  is  an  arch,  as  the  angle  ABC  is  to  four  right 
angles;  therefore  the  arch  AC  is  to  the  whole  circumference  of 
which  it  is  an  arch,  as  the  arch  DE  to  the  whole  circumference 
of  which  it  is  an  arch. 

DEFINITIONS.    FIG.  3. 

I. 

Let  ABC  be  a  plane  rectilineal  angle;  if  about  B  as  a  centre, 
with  BA  any  distance,  a  circle  ACF  be  described,  meeting  BA, 
BC  in  A,  C;  the  arch  AC  is  called  the  measure  of  the  angle 
ABC. 

II. 

The  circumference  of  a  circle  is  supposed  to  be  divided  into 
360  equal  parts  called  degrees;  and  each  degree  into  60  equal 
parts  called  minutes,  and  each  minute  into  60  equal  parts  called 
seconds,  &c.  And  as  many  degrees,  minutes,  seconds,  &c.  as 
are  contained  in  any  arch,  of  so  many  degrees,  minutes,  se- 
conds, Sec.  is  the  angle,  of  which  that  arch  is  the  measure,  said 
to  be. 

Cor.  Whatever  be  the  radius  of  the  circle  of  which  the  measure 
of  a  given  angle  is  an  arch,  that  arch  will  contain  the  same 
number  of  degrees,  minutes,  seconds,  8cc.  as  is  manifest  from 
Lemma  2. 


388  PLANE  TRlGONOMETRYo 

III. 

Let  AB  be  produced  till  it  meet  the  circle  again  in  F;  the  angle 
CBF,  which,  together  with  ABC,  is  equal  to  two  right  angles, 
is  called  the  Supplement  of  the  angle  ABC. 

IV. 

A  straight  line  CD  drawn  through  C,  one  of  the  extremities  of  the 
arch  AC  perpendicular  upon  the  diameter  passing  through  the 
other  extremity  A,  is  called  the  Sine  of  the  arch  AC,  or  of  the 
angle  ABC,  of  which  it  is  the  measure. 

CoR.  The  Sine  of  a  quadrant,  or  of  a  right  angle,  is  equal  to  the 
radius. 

V. 

The  segment  DA  of  the  diameter  passing  through  A,  one  extre- 
mity of  the  arch  AC,  between  the  sine  CD  and  that  extremity^ 
is  called  the  Versed  Sine  of  the  arch  AC,  or  angle  ABC. 

VI. 

A  straight  line  AE,  touching  the  circle  at  A,  one  extremity  of 
the  arch  AC,  and  meeting  the  diameter  BC  passing  through  the 
other  extremity  C  in  E,  is  called  the  Tangent  of  the  arch  AC; 
or  of  the  angle  ABC. 

VII. 

The  straight  line  BE,  between  the  centre  and  the  extremity  of 
the  tangent  AE,  is  called  the  Secant  of  the  arch  AC,  or  angle 
ABC. 

Cor.  to  def.  4,  6,  7.  The  sine,  tangent,  and  secant  of  any  angle 
ABC,  are  likewise  the  sine,  tangent,  and  secant  of  its  supple- 
ment CBF. 

It  is  manifest  from  def.  4,  that  CD  is  the  sine  of  the  angle  CBF. 
Let  CB  be  produced  till  it  meet  the  circle  again  in  G;  and  it  is 
manifest  that  AE  is  the  tangent,  and  BE  the  secant,  of  the  angle 
ABG  or  EBF,  from  def.  6,  7. 

Cor.  to  def.  4,  5,  6,  7.  The  sine,  versed  sine,  tangent,  and  secant, 
of  any  arch  which  is  the  measure  of  any  given  angle  ABC,  is  to 
the  sine,  versed  sine,  tangent,  and  secant,  of  any  other  arch 
which  is  the  measure  of  the  same  angle,  as  the  radius  of  the  first 
is  to  the  radius  of  the  second. 

Let  AC,  MN  be  measures  of  the  angle  ABC,  according  to  def.  1, 
CD  the  sine,  DA  the  versed  sine,  AE  the  tangent,  and  BE  the 
secant  of  the  arch  AC,  according  to  def.  4,  5,  6,  7,  and  NO  the 
sine,  OM  the  versed  sine,  MP  the  tangent,  and  BP  the  secant 
of  the  arch  MN,  according  to  the  same  definitions.  Since  CD, 
NO,  AE,  MP  are  parallel,  CD  is  to  NO  as  the  radius  CB  to 
the  radius  NB,  and  AE  to  MP  as  AB  to  BM,  and  BC  or  BA  to 
BD  as  BN  or  BM  to  BO;  and,  by  conversion,  DA  to  MO  as  4y 
AB  to  MB.  Hence  the  corollary  is  manifest;  therefore,  if  the 
radius  be  supposed  to  be  divided  into  any  given  number  of 
equal  parts,  the  sine,  versed  sine,  tangent,  and  secant,  of  any 


PLANE  TRIGONOMETRY.  389 

given  angle,  will  each  contain  a  given  number  of  these  parts; 
and,  by  trigonometrical  tables,  the  length  of  the  sine,  versed 
sine,  tangent,  and  secant,  of  any  angle,  may  be  found  in  parts  of 
which  the  radius  contains  a  given  number;  and,  vice  versa,  a 
number  expressing  the  length  of  the  sine,  versed  sine,  tangent, 
and  secant,  being  given,  the  angle  of  which  it  is  the  sine,  versed 
sine,  tangent,  and  secant,  may  be  found. 


VIII.  Fig.  3. 

The  difference  of  an  angle  from  a  right  angle  is  called  the  com- 
plement oi  that  angle.  Thus,  if  BH  be  drawn  perpendicular  to 
AB,  the  angle  CBH  will  be  the  complement  of  the  angle  ABC, 
or  of  CBF. 

IX. 

Let  HK  be  the  tangent,  CL  or  DB,  which  is  equal  to  it,  the  sine 
and  BK  the  secant  of  CBH,  the  complement  of  ABC,  according 
to  def.  4,  6,  7,  HK  is  called  the  co-tangent,  BD  the  co-sine,  and 
BK  the  cosecant,  of  the  angle  ABC. 

Cor.  1.  The  radius  is  a  mean  proportional  between  the  tangent 
and  co-tangent. 

For,  since  HK,  BA  are  parallel,  the  angles  HKB,  ABC  will  be 
equal,  and  the  angles  KHB,  BAE  are  right:  therefore  the  tri- 
angles BAE,  KHB  are  similar,  and  therefore  AE  is  to  AB,  as 
BH  or  BA  to  HK. 

Cor.  2.  The  radius  is  a  mean  proportional  between  the  co-sine 
and  secant  of  any  angle  ABC. 

Since  CD,  AE  are  parallel,  BD  is  to  BC  or  BA,  as  BA  to  BE. 

PROP.  I.   FIG.  5. 

In  a  right  angled  plane  triangle,  if  the  hypothenuse 
be  made  radius,  the  sides  become  the  sines  of  the  angles 
opposite  to  them:  and  if  either  side  be  made  radius^ 
the  remaining  side  is  the  tangent  of  the  angle  opposite 
to  it,  and  the  hypothenuse  the  secant  of  the  same  angle. 

Let  ABC  be  a  right  anglc^d  triangle;  if  the  hypothenuse  BC  be 
made  radius,  either  of  the  sides  AC  will  be  the  sine  of  the  angle 
ABC  opposite  to  it;  and  if  either  side  BA  be  made  radius,  the 
other  side  AC  will  be  the  tangent  of  the  angle  ABC  opposite  to  it> 
and  the  hypothenuse  BC  the  secant  of  the  same  angle. 

About  B  as  a  centre,  with  BC,  BA  for  distances,  let  two  circles 
CD,  EA  be  described  meeting  BA,  BC  in  D,  E:  since  CAB  is  a 
right  angle,  BC  being  radius,  AC  is  the  sine  of  the  angle  ABC  by 
def.  4,  and  BA  being  radius,  AC  is  the  tangent,  and  BC  the  secant 
of  the  angle  ABC,  by  def.  6,  7. 

Cor.  1.  Of  the  hyjiothenuse  a  side  and  an  angle  of  a  right  an- 
gled triangle,  any  two  being  given,  the  third  is  also  given. 

Cor.  2.  Of  the  two  sides  and  an  angle  of  a  right  angled  trian- 
gle, any  two  being  given,  the  third  is  also  given. 


390  PLANE  TRIGONOMETRY. 

PROP.  II.    FIG.  6,  7. 

The  sides  of  a  plane  triangle  are  to  one  another  as 
the  sines  of  the  angles  opposite  to  them. 

In  right  angled  triangles,  this  prop,  is  manifest  from  prop.  1; 
for  if  the  hypothenuse  be  made  radius,  the  sides  are  the  sines  of 
the  angles  opposite  to  them,  and  the  radius  is  the  sine  of  a  right 
angle  (cor.  to  def.  4.)  which  is  opposite  to  the  hypothenuse. 

In  any  oblique  angled  triangle  ABC,  any  two  sides  AB,  AC  will 
be  to  one  another  as  the  sines  of  the  angles  ACB,  ABC  which  are 
opposite  to  them. 

From  C,  B  draw  CE,  BD  perpendicular  upon  the  opposite  sides 
AB,  AC  produced,  if  need  be.  Since  CEB,  CDB  are  right  angles, 
BC  being  radius,  CE  is  the  sine  of  the  angle  CBA,  and  BD  the 
sine  of  the  angle  ACB:  but  the  two  triangles  CAE,  DAB  have 
each  a  right  angle  at  D  and  E;  and  likewise  the  common  angle 
CAB;  therefore  they  are  similar,  and  consequently,  CA  is  to  AB, 
as  CE  to  DB;  that  is,  the  sides  are  as  the  sines  of  the  angles 
opposite  to  them. 

Cor.  Hence  of  two  sides,  and  two  angles  opposite  to  them,  in  a 
plane  triangle,  any  three  being  given,  the  fourth  is  also  given. 

PROP.  III.    FIG.  8. 

In  a  plane  triangle,  the  sum  of  any  two  sides  is  to 
their  difference,  as  the  tangent  of  half  the  sum  of  the 
angles  at  the  base,  to  the  tangent  of  half  their  differ- 
ence. 

Let  ABC  be  a  plane  triangle;  the  sum  of  any  two  sides,  AB,  AC 
will  be  to  their  difference  as  the  tangent  of  half  the  sum  of  the  an- 
gles at  the  base  ABC,  ACB  to  the  tangent  of  half  their  difference. 

About  A  as  a  centre,  with  AB  the  greater  side  for  a  distance, 
let  a  circle  be  described,  meeting  AC  produced  in  E,  F,  and  BC 
in  D;  join  DA,  EB,  FB:  and  draw  FG  parallel  to  BC,  meeting  EB 
in  G. 

The  angle  EAB  (32.  1.)  is  equal  to  the  sum  of  the  angles  at  the 
base,  and  the  angle  EFB  at  the  circumference  is  equal  to  the  half 
of  EAB  at  the  centre  (20.  3.);  therefore  EFB  is  half  the  sum  of  the 
angles  at  the  base;  but  the  angle  ACB  (32.  1.)  is  equal  to  the  an- 
gles CAD  and  ADC,  or  ABC  together:  therefore  FAD  is  the  dif- 
ference of  the  angles  at  the  base,  and  FBD  at  the  circumference, 
or  BFG,  on  account  of  the  parallels  FG,  BD,  is  the  half  of  that 
difference;  but  since  the  angle  EBF  in  a  semicircle  is  a  right  an- 
gle (l.  of  this),  FB  being  radius,  BE,  BG,  are  the  tangents  of  the 
angles  EFB,  BFG;  but  it  is  manifest  that  EC  is  the  sum  of  the 
sides  BA,  AC,  and  CF  their  difference;  and  since  BC,FG  are  pa- 
rallel (2.  6.),  EC  is  to  CF,  as  EB  to  BG;  that  is,  the  sum  of  the 
sides  is  to  their  difference,  as  the  tangent  of  half  the  sum  of  the 
angles  at  the  base  to  the  tangent  of  half  their  difference. 


PLANE    TRIGONOMETRY.  391 


PROP.  IV.    FIG.  18. 


In  any  plane  triangle  BAG,  whose  two  sides  are  BA, 
AC,  and  base  BC,  the  less  of  the  two  sides,  which  let  be 
BA,  is  to  the  greater  AC,  as  the  radius  is  to  the  tan- 
gent of  an  angle;  and  the  radius  is  to  the  tangent  of 
the  excess  of  this  angle  above  half  a  right  angle,  as  the 
tangent  of  half  the  sum  of  the  angles  B  and  C  at  the 
base  is  to  the  tangent  of  half  their  difference. 

At  the  point  A,  draw  the  straight  line  EAD  perpendicular  to 
BA:  make  AE,  AF  each  equal  to  AB,  and  AD  to  AC;  join  BE, 
BF,  BD,  and  from  D  draw  DG  perpendicular  upon  BF.     And  be-- 
cause  BA  is  at  right  angles  to  EF,  and  EA,  AB,  AF  are  equal, 
each  of  the  angles  EBA,  ABF  is  half  a  right  angle,  and  the  whole 
EBF  is  a  right  angle;  also  (4.  1.  El.)  EB  is  equal  to  BF.     And 
since  EBF,  FGD  are  right  angles,  EB  is  parallel  to  GD,  and  the 
triangles  EBF,  FGD  are  similar;  therefore  EB  is  to  BF  as  DG  to 
GF,  and  EB  being  equal  to  BF,  FG  must  be  equal  to  GD.     And 
because  BAD  is  a  right  angle,  BA  the  less  side  is  to  AD  or  AC 
the  greater,  as  the  radius  is  to  the  tangent  of  the  angle  ABD;  and 
because  BGD  is  a  right  angle,  BG  is  to  GD  or  GF  as  the  radius  is 
to  the  tangent  of  GBD,  which  is  the  excess  of  the  angle  ABD 
above  ABF  half  a  right  angle.    But  because  EB  is  parallel  to  GD, 
BG  is  to  GF  as  ED  is  to  DF,  that  is,  since  ED  is  the  sum  of  the 
sides  BA,  AC,  and  FD  their  difference  (3.  of  this),  as  the  tangent 
of  half  the  sum  of  the  angles  B,  C,  at  the  base,  to  the  tangent  of 
half  their  difference.    Therefore,  in  any  plane  triangle,  Sec.  Q.  E.  D. 

PROP.  V.     FIG.  9.  10. 

In  any  triangle,  twice  the  rectangle  contained  by  any 
two  sides  is  to  the  difference  of  the  sum  of  the  squares 
of  these  two  sides,  and  the  square  of  the  base,  as  the 
radius  is  to  the  co-sine  of  the  angle  included  by  the 
two  sides. 

Let  ABC  be  a  plane  triangle;  twice  the  rectangle  ABC  contained 
by  any  two  sides  BA,  BC,  is  to  the  difference  of  the  sum  of  the 
squares  of  BA,  BC,  and  the  square  of  the  base  AC,  as  the  radius 
to  the  co-sine  of  the  angle  ABC. 

From  A,  draw  AD  perpendicular  upon  the  opposite  side  BC; 
then  (by  12.  and  13.  2.  El.)  the  difference  of  the  sum  of  the  squares 
of  AB,  BC,  and  the  square  of  the  base  AC,  is  equal  to  twice  the 
rectangle  CBD;  but  twice  the  rectangle  CBA  is  to  twice  the  rect- 
angle CBD,  that  is  to  the  difference  of  the  sum  of  the  squares  of 
AB,  BC,  and  the  square  of  AC  (l.  6.),  as  AB  to  BD;  that  is,  by 
prop.  1,  as  radius  to  the  sine  of  BAD,  which  is  the  complement  of 
the  angle  ABC,  that  is,  as  radius  to  the  co-sine  of  ABC. 


392  PLANE    TRIGONOMETRY. 


PROP.  VI.     FIG.  11. 


In  any  triangle  ABC,  whose  two  sides  are  AB,  AC, 
and  base  BC,  the  rectangle  contained  by  half  the  pe- 
rimeter, and  the  excess  of  it  above  the  base  BC,  is  to 
the  rectangle  contained  by  the  straight  lines  by  which 
the  half  of  the  perimeter  exceeds  the  other  two  sides 
AB,  AC,  as  the  square  of  the  radius  is  to  the  square  of 
the  tangent  of  half  the  angle  B  AC  opposite  to  the  base. 

Let  the  angles  BAG,  ABC  be  bisected  by  the  straight  lines  AG, 
BG;  and  producing  the  side  AB,  let  the  exterior  angle  CBH  be 
bisected  by  the  straight  line  BK,  meeting  AG  in  Kj  and  from  the 
points  G,  K,  let  there  be  drawn  perpendicular  upon  the  sides  the 
straight  lines  GD,  GE,  GF,  KH,  KL,  KM.    Since  therefore  (4.  4.) 
G  in  the  centre  of  the  circle  inscribed  in  the  triangle  ABC,  QT>, 
GF,  GE  will  be  equal,  and  AD  will  be  equal  to  AE,  BD  to  BF,  and 
CE  to  CF;  in  like  manner  KH,  KL,  KM  will  be  equal,  and  BH 
will  be  equal  to  BM,  and  AH  to  AL,  because  the  angles  HBM, 
HAL  are  bisected  by  the  straight  lines  BK,  KA:  and  because  in 
the  triangles  KCL,  KCM,  the  sides  LK,  KM  are  equal,  KC  is  com- 
mon, and  KLC,  KMC  are  right  angles,  CL  will  be  equal  to  CM: 
since  therefore  BM  is  equal  to  BH,  and  CM  to  CL^  BC  will  be 
equal  to  BH  and  CL  together^  and,  adding  AB  and  AC  together, 
AB,  AC,  and  BC  will  together  be  equal  to  AH  and  AL  together: 
but  AH,  AL  are  equal:  wherefore  each  of  them  is  equal  to  half 
the  perimeter  of  the  triangle  ABC:  but  since  AD,  AE  are  equal, 
and  BD,  BF,  and  also  CE,  CF,  AB,  together  with  FC,  will  be  equal 
to  half  the  perimeter  of  the  triangle   to  which  AH  or  AL  was 
shown  to  be  equalj  taking  away  therefore  the  common  AB,  the  re- 
mainder FC  will  be  equal  to  the  remainder  BH;  in  the  same  man- 
ner it  is  demonstrated,  that  BF  is  equal  to  CL;  and  since  the 
points  B,  D,  G,  F,  are  in  a  circle,  the  angle  DGF  will  be  equal  to 
the  exterior  and   opposite  angle  FBH  (22.  3.);  wherefore  their 
halves  BGD,  HBK  will  be  equal  to  one  another:  the  right  angled 
triangles  BGD,  HBK  will  therefore  be  equiangular,  and  GD  will 
be  to  BD,  as  BH  to  HK,  and  the  rectangle  contained  by  GD,  HK 
will  be  equal  to  the  rectangle  DBH  or  BFC;  but  since  AH  is  to 
HK,  as  AD  to  DG,  the  rectangle  HAD  (22.  6.)  will  be  to  the  rect- 
angle contained  by  HK,  DG,  or  the  rectangle  BFC,  (as  the  square 
of  AD  is  to  the  square  of  DG,  that  is)  as  the  square  of  the  radius 
to  the  square  of  the  tangent  of  the  angle  DAG,  that  is,  the  half  of 
BAC:  but  HA  is  half  the  perimeter  of  the  triangle  ABC,  and  AD 
is  the  excess  of  the  same  above  HD,  that  is,  above  the  base  BC: 
but  BF  or  CL  is  the  excess  of  HA  or  AL  above  the  side  AC;  and 
FC,  or  HBjis  the  excess  of  the  same  HA  above  the  side  AB;  there- 
fore the  rectangle  contained  by  half  the  perimeter,  and  the  excess 
of  the  same  above  the  base,  viz.  the  rectangle  HAD,  is  to  the  rect- 
angle contained  by  the  straight  lines  by  which  the  half  of  the  peri- 
meter exceeds  the  other  two  sides,  that  is,  the  rectangle  BFC,  as 


PLANE   TRIGONOMETRY.  393 

the  square  of  the  radius  is  to  the  square  of  the  tangent  of  half  the 
angle  BAG  opposite  to  the  base.   Q.  E.  D. 

PROP.  VII.     FIG.  12.   13. 

In  a  plane  triangle,  the  base  is  to  the  sum  of  the  sides? 
as  the  difference  of  the  sides  is  to  the  sum  or  difference 
of  the  segments  of  the  base  made  by  the  perpendicular 
upon  it  from  the  vertex,  according  as  the  square  of  the 
greater  side  is  greater  or  less  than  the  sum  of  the 
squares  of  the  lesser  side  and  the  base. 

Let  ABC  be  a  plane  triangle;  if  from  A  the  vertei  be  drawn  a 
straight  line  AD  perpendicular  upon  the  base  BC,  the  base  BC 
will  be  to  the  sum  of  the  sides  BA,  AC,  as  the  difference  of  the 
same  sides  is  to  the  sum  or  difference  of  the  segments  CD,  BD, 
according  as  the  square  of  AC  the  greater  side  is  greater  or  less 
than  the  sum  of  the  squares  of  the  lesser  side  AB,  and  the  base 
BC. 

About  A  as  a  centre,  with  AC  the  greater  side  for  a  distance, 
let  a  circle  be  described  meeting  AB  produced  in  E,  F,  and  CB 
in  G:  it  is  manifest,  that  FB  is  the  sum,  and  BE  the  difference  of 
the  sides;  and  since  AD  is  perpendicular  to  GC,  GD,  CD  will  be 
equal;  consequently  GB  will  be  equal  to  the  suin  or  difference  of 
the  segments  CD,  BD,  according  as  the  perpendicular  AD  meets 
the  base,  or  the  base  produced;  that  is,  (by  conv.  12.  13.  2.)  ac- 
cording as  the  square  of  AC  is  greater  or  less  than  the  sum  of  the 
squares  of  AB,  BC:  but  (by  35.  3.)  the  rectangle  CBG  is  equal  to 
the  rectangle  EBF;  that  is,  (16.  6.)  BC  is  to  BF,  as  BE  is  to  BG, 
that  is,  the  base  is  to  the  sum  of  the  sides,  as  the  difference  of  the 
sides  is  the  sum  or  diff"erence  of  the  segments  of  the  base  made  by 
the  perpendicular  from  the  vertex,  according  as  the  square  of  the 
greater  side  is  greater  or  less  than  the  sum  of  the  squares  of  the 
lesser  side  and  the  base.     Q.  E.  D, 

PROP.  VIII.     PROB.     FIG.  14. 

The  sum  and  difference  of  two  magnitudes  beino; 
given,  to  find  them. 

Half  the  given  sum  added  to  lialf  the  given  difference,  will  be 
the  greater,  and  half  the  difference  subtracted  from  half  the  sum, 
will  be  the  less.  *, 

For,  let  AB  be  the  given  sum,  AC  the  greater,  and  BC  the  less. 
Let  AD  be  half  the  given  sum;  and  to  AD,  DB,  which  are  equal, 
let  DC  be  added;  then  AC  will  be  equal  to  BD  and  DC  together; 
that  is,  to  BC,  and  twice  DC;  consequently  twice  DC  is  the  dif- 
ference, and  DC  half  that  difference;  but  AC  the  greater  is  equal 
to  AD,  DC;  that  is,  to  half  the  sum  added  to  half  the  diff"erence, 
and  BC  the  less  is  equal  to  the  excess  of  BD,  half  the  sum  above 
DC  half  the  difference.  Q.  E.  D. 
3D 


394  PLANE  TRIGONOMETRY. 

SCHOLIUM. 

Of  the  six  parts  of  a  plane  triangle  (the  three  sides  and  three 
angles)  any  three  being  given,  to  find  the  other  three  is  the  busi- 
ness of  plane  trigonometry^  and  the  several  cases  of  that  problem 
may  be  resolved  by  means  of  the  preceding  proposition,  as  in  the 
two  following,  with  the  tables  annexed.  In  these,  the  solution  is 
expressed  by  a  fourth  proportional  to  three  given  lines;  but  if  the 
given  parts  be  expressed  by  numbers  from  trigonometrical  ta- 
bles, it  may  be  obtained  arithmetically  by  the  common  Rule  of 
Three. 


Note.  In  the  tables,  the  following  abbreviations  are  used :  R  is  put  for  the 
Radius;  T  for  Tangent;  and  S  for  Sine.  Degrees,  minutes,  seconds,  &c.  are 
written  in  this  manner ;  30°  25'  13",  &c.,  which  signifies  30  degrees,  25  mi- 
nutes, 13  seconds,  <&c. 


PLANE  TRIGONOMETRY. 


395 


SOLUTION  OF  THE  CASES  OF  RIGHT  ANGLED  TRIANGLES. 

GENERAL  PROPOSITION. 

In  a  right  angled  triangle,  of  the  three  sides  and 
three  angles,  any  two  being  given  besides  the  right 
angle,  the  other  three  may  be  found,  except  when  the 
two  acute  angles  are  given;  in  which  case  the  ratios 
of  the  sides  are  only  given,  being  the  same  with  the 
ratios  of  the  sines  of  the  angles  opposite  to  them. 

It  is  manifest  from  47,  1,  that  if  of  the  two  sides  and  hypothe- 
nuse  any  two  be  given,  the  third  may  also  be  found.  It  is  also 
manifest  from  32,  1,  that  if  one  of  the  acute  angles  of  a  right  an- 
gled triangle  be  given,  the  other  is  also  given,  for  it  is  the  com- 
plement of  the  former  to  a  right  angle. 

If  two  angles  of  any  triangle  be  given,  the  third  is  also  given, 
being  the  supplement  of  the  two  given  angles  to  two  right  an- 
gles. 

The  other  cases  may  be  resolved  by  help  of  the  preceding  pro- 
positions, as  in  the  following  table: 


GIVEN. 


SOUGHT. 


1 


Two     sides,    AB, 
AC. 


AB,BC,asideand 
the  hypothenuse. 


AB,  B,  a  side  and 
an  angle. 


AB  and  B,  a  side 
and  an  angle. 


BC  and  B,  the  hy- 
pothenuse and  an  an 
gle. 


The    an- 
gles B,  C. 


The    an- 
gles B,  C. 


The  other 
side  AC. 


The    hy- 
pothenuse 
BC. 


The  side 
AC. 


AB  :  AC  :  :  R  :  T,  B,  of 

which  C  is  the  complement. 


BC  :  BA   :  :  R  :  S,  C,  of 

which  B  is  the  complement. 


R  :  T,  B  :  :  BA  :  AC. 


S,  C  :  R  :  :  BA  :  BC. 


R 


S,  B 


BC  :  CA. 


These  five  cases  are  resolved  by  prop.  1, 


396 


PLANE  TRIGONOMETRY, 


SOLUTION  OF  THE  GASES  OF  OBLIQUE  ANGLED  TRIANGLES. 

GENERAL  PROPOSITION. 

In  an  oblique  angled  triangle,  of  the  three  sides  and 
three  angles,  any  three  being  given,  the  other  three 
may  be  found,  except  when  the  three  angles  are  given; 
in  which  case  the  ratios  of  the  sides  are  only  given, 
being  the  same  with  the  ratios  of  the  sines  of  the  an- 
gles opposite  to  them. 


GIVEN. 


SOUGHT. 


A,  B,  and  there- 
fore C,  and  the  side 
AB. 


AB,  AC,  and  B, 

two  sides  and  an 
angle  opposite  to 
one  of  them. 


AB,  AC,  and  A, 
two  sides  and  the 
included  angle. 


BC,  AC. 


The 
gles  A 
C. 


an- 
and 


The  an- 
gles B  and 
C. 


S,  C  :  S,  A  : :  AB  :  BC,  and 
also,  S,  C  :  S,  B  : :  AB  :  AC. 

(2.)     (Fig.  16.  17.) 


AC:  AB  ::  S,B  :  S,C,  (2.) 
This  case  admits  of  two  solu 
tions;  for  C  may  be  greatei 
or  less  than  a  quadrant.  (Cor. 
to  def.  4.) 


AB+AC  :  AB— AC  : :  T, 
C+B  :  T,C— B  :  (3.)  the  sum 


2        ,  2 

and  difference  of  the  angles  C, 
B  being  given  each  of  them  is 
given.  (7.)  Otherwise.  Fig.  1 8. 
BA  :  AC  : :  R  :  T,  ABC 
and  also  R  :  T,  ABC— 45° 
T,B+C  ;  T,  B— C  ;  (4.)there- 

2  2 

fore  B  and  C  are  given  as  be- 
fore. (7.) 


PLANE  TRIGONOMETRY. 


397 


GIVEN. 


SOUGHT. 


4 


AB,  BC,  CA,  the 
three  sides. 


A,  B,  C, 

the     three 
angles. 


2  AC  X  CB  :  AC^  +  CB^r— 
ABq  :  :  R  :  Co  S,  C.  If  ABq 
+  CB^r  be  greater  than  ABq. 
Fig.  16. 

2    AC  X  CB  :    ABq^-ACq— 
CB^+  ::  R  :  CoS,C.     If  AB7 
be  greater  than  AC</xCBgf.  Fig 
17.  (4.) 

Otherwise. 

Let  AB+BC+AC=2  P.  P+ 


P—AB  :  P~AC  +  P— BC   : 
Rq  :  Tq.  5   C,  and  hence  C  is 
known.  (5.) 

Otherwise. 

Let  AD  be  perpendicular  to 
BC.  1.  If  ABq  be  less  than 
ACq+CBq.  Fig.  16.  BC  :  BA 
+AC  :  :  BA— AC  :  BD— DC, 
and  BC  the  sum  of  BD,  DC  is 
given^  therefore  each  of  them 
is  given.  (7.) 

2.  If  ABq  be  greater  than 
AC^+CB^.  Fig.  17.  BC  :  BA 
+AC  :  :  BA— AC  :  BD+DC; 
and  BC  the  difference  of  BD, 
DC  is  given,  therefore  each  of 
them  is  given.  (7.) 

And  CA  :  CD  :  :  R  :  Co  S, 
C.  ( 1 .)  and  C  being  found,  A  and 
B  are  found  by  case  2,  or  3. 


*■ 


SPHERICAL  TRIGONOMETRY. 


DEFINITJONS. 

I. 

The  pole  of  a  circle  of  the  sphere  is  a  point  in  the  superficies  of 
the  sphere,  from  which  all  straight  lines  drawn  to  the  circum- 
ference of  the  circle  are  equal. 

II. 

A  great  circle  of  the  sphere  is  any  whose  plane  passes  through 
the  centre  of  the  sphere,  and  whose  centre  therefore  is  the  same 
with  that  of  the  sphere. 

III. 
A  spherical  triangle  is  a  figure  upon  the  superficies  of  a  sphere 
comprehended  by  three  arches  of  three  great  circles,  each  of 
which  is  less  than  a  semicircle. 

IV. 

A  spherical  angle  is  that  which  on  the  superficies  of  a  sphere  is 
contained  by  two  arches  of  great  circles,  and  is  the  same  with 
the  inclination  of  the  planes  of  these  great  circles. 

PROP.  I. 
Great  circles  bisect  one  another. 

As  they  have  a  common  centre,  their  common  section  will  be 
a  diameter  of  each  which  will  bisect  them. 

PROP.  II.    FIG.  1. 

The  arch  of  a  great  circle  betwixt  the  pole  and  the 
circumference  of  another  is  a  quadrant. 

Let  ABC  be  a  great  circle,  and  D  its  pole;  if  a  great  circle 
DC  pass  through  D,  and  meet  ABC  in  C,  the  arch  DC  will  be  a 
quadrant. 

Let  the  great  circle  CD  meet  ABC  again  in  A,  and  let  AC  be 
the  common  section  of  the  great  circles  which  will  pass  through 
E  the  centre  of  the  sphere:  join  DE,  DA,  DC:  by  def.  1,  DA, 
DC  are  equal,  and  AE,  EC  are  also  equal,  and  DE  is  commonj 
therefore  (8.  1.)  the  angles  DEA,  DEC  are  equal;  wherefore  the 
arches  DA,  DC  are  equal,  and  consequently  each  of  them  is  a 
quadrant.    Q.  E.  D. 

PROP.  III.    FIG.  2. 

If  a  great  circle  be  described  meeting  two  great  cir- 
cles AB,  AC  passing  through  its  pole  A  in  B,  C,  the  an- 
gles at  the  centre  of  the  sphere  upon  the  circumference 


SPHERICAL    TRIGONOMETRY.  399 

BC,  is  the  same  with  the  spherical  angle  BAC,  and  the 
arch  BC  is  called  the  measure  of  the  spherical  angle 
BAC. 

Let  the  planes  of  the  great  circles  AB,  AC  intersect  one  ano- 
ther in  the  straight  line  AD  passing  through  D  their  common 
centre^  join  DB,  jDC. 

Since  A  is  the  pole  of  BC,  AB,  AC  will  be  quadrants,  and  the 
angles  ADB,  ADC  right  angles;  therefore  (6.  def.  1 1.),  the  angle 
CDB  is  the  inclination  of  the  planes  of  the  circles  AB,  AC:  that 
is,  (def.  4.)  the  spherical  angle  BAC.     Q.  E.  D. 

CoR.  If  through  the  point  A,  two  quadrants  AB,  AC  be  drawn, 
the  point  A  will  be  the  pole  of  the  great  circle  BC,  passing 
through  their  extremities  B,  C. 

Join  AC,  and  draw  AE  a  straight  line  to  any  other  point  E  in 
BC;  join  DE:  since  AC,  AB  are  quadrants,  the  angles  ADB, 
ADC  are  right  angles,  and  AD  will  be  perpendicular  to  the  plane 
of  BC:  therefore  the  angle  ADE  is  a  right  angle,  and  AD,  DC 
are  equal  to  AD,  DE,  each  to  each;  therefore  AE,  AC  are  equal, 
and  A  is  the  pole,  of  BC,  by  def.  i.     Q.  E.  D. 

PROP.  IV.     FIG.  3. 

In  isosceles  spherical  triangles,  the  angles  at  the  base 
are  equal. 

Let  ABC  be  an  isosceles  triangle,  and  AC,  CB  the  equal  sides; 
the  angles  BiVC,  ABC,  at  the  base  AC,  are  equal. 

Let  D  be  the  centre  of  the  sphere,  and  join  DA,  DB,  DC;  in 
DA  take  any  point  E,  from  which  draw,  in  the  plane  ADC,  the 
straight  line  EF  at  right  angles  to  ED,  meeting  CD  in  F,  and 
draw,  in  the  plane  ADB,  EG  at  right  angles  to  the  same  ED; 
therefore  the  rectilineal  figure  FEG  is  (6.  def.  1 1.)  the  inclination 
of  the  planes  ADC,  ADB,  and  therefore  is  the  same  with  the 
spherical  angle  BAC:  from  F  draw  FH  perpendicular  to  DB, 
and  from  H  draw,  in  the  plane  ADB,  the  straight  line  HG  at 
right  angles  to  HD,  meeting  EG  in  G,  and  join  GF.  Because 
DE  is  at  right  angles  to  EF  and  EG,  it  is  perpendicular  to  the 
plane  FEG  (4.  1 1.),  and  therefore  the  plane  FEG,  is  perpendicu- 
lar to  the  plane  ADB,  in  which  DE  is  (18.  11.):  in  the  same  man- 
ner the  plane  FHG  is  perpendicular  to  the  plane  ADB,  and  there- 
fore GF  the  common  section  of  the  planes  FEG,  FHG,  is  per- 
pendicular to  the  plane  ADB  (19.  11.);  and  because  the  angle 
FHG  is  the  inclination  of  the  planes  BDC,  BDA,  it  is  the  same 
with  the  spherical  angle  ABC;  and  the  sides  AC,  CB  of  the  sphe- 
rical triangle  being  equal,  the  angles  EDF,  HDF,  which  stand 
upon  them  at  the  centre  of  the  sphere,  are  equal;  and  in  the  tri- 
angles EDF,  HDF,  the  side  DF  is  common,  and  the  angles  DEF, 
DHF  are  right  angles;  therefore  EF,  FPI  are  equal,  and  in  the 
triangles  FEG,  FHG  the  side  GF  is  common,  and  the  sides  EG, 
GH  will  be  equal  by  the  47.  1.  and  therefore  the  angle  FEG  is 
equal  to  FHG  (8.  1.);  that  is,  the  spherical  angle  BAC  is  equal 
to  the  spherical  angle  ABC, 


400  SPHERICAL   TRIGONOMETRY. 

PROP.  V.    FIG.  3. 

If,  in  a  spherical  triangle  ABC,  two  of  the  angles 
BAG,  ABC,  be  equal,  the  sides  BC,  AC,  opposite  to 
them,  are  equal. 

Read  the  construction  and  demonstra,tion  of  the  preceding  pro- 
position, unto  the  words  "and  the  sides  AC,  CB,'*  &c.  and  the 
rest  of  the  demonstration  will  be  as  follows,  viz. 

And  the  spherical  angles  BAG,  ABC,  being  equal,  the  recti- 
lineal angles  FEG,  FHG,  which  are  the  same  with  them,  are 
equalj  and  in  the  triangles  FGE,  FGH  the  angles  at  G  are  right 
angles,  and  the  side  FG  opposite  to  two  of  the  equal  angles  is 
common j  therefore  (26.  1.)  EF  is  equal  to  FH^  and  in  the  right 
angled  triangles  DEF,  DHF  the  side  1)F  is  common^  wherefore 
(47.  1.)  ED  is  equal  to  DH,  and  the  angles  EDF,  HDF,  are  there- 
fore equal  (4.  1.),  and  consequently  the  sides  AC,  BC  of  the  sphe- 
rical triangle  are  equal. 

PROP.  VI.    FIG.  4. 

Any  two  sides  of  a  spherical  triangle  are  greater  than 
the  third. 

Let  ABC  be  a  spherical  triangle,  any  two  sides  AB,  BC  will 
be  greater  than^  the  other  side  AC. 

Let  D  be  the  centre  of  the  sphere;  join  DA,  DB,  DC. 

The  solid  angle  at  D,  is  contained  by  three  plane  angles  ADB, 
ADC,  BDC;  and,  by  20.  11.  any  two  of  them  ADB,  BDC  are 
greater  than  the  third  ADC;  that  is,  any  two  sides  AB,  BC  of 
the  spherical  triangle  ABC,  are  greater  than  the  third  AC. 

PROP.  VII.     FIG.  4. 

The  three  sides  of  a  spherical  triangle  are  less  than  r 
circle. 

Let  ABC  be  a  spherical  triangle  as  before,  the  three  sides  AB, 
BC,  AC,  are  less  than  a  circle. 

Let  D  be  the  centre  of  the  sphere:  the  solid  angle  at  D  is  con- 
tained by  three  plane  angles  BDA,  BDC,  ADC,  which  together 
are  less  than  four  right  angles  (21.  11.);  therefore  the  sides  AB, 

BC,  AC  together,  will  be  less  than  four  quadrants:  that  is,  less 
than  a  circle. 

PROP.  VIII.    FIG.  5. 

In  a  spherical  triangle  the  greater  angle  is  opposite 
to  the  greater  side;  and  conversely. 

Let  ABC  be  a  spherical  triangle,  the  greater  angle  A  is  opposed 
to  the  greater  side  BC. 

Let  the  angle  BAD  be  made  equal  to  the  angle  B,  and  then 

BD,  DA  will  be  equal,  (5.  of  this,)  and  therefore  AD,  DC  are 
equal  to  BC,  but  AD,  DC,  are  greater  than  AC,  (6.  of  this), 
therefore  BC  is  greater  than  AC,  that  is,  the  greater  angle  A  is 


SPHERICAL  TRIGONOMETRY.  401 

opposite  to  the  greater  side  BC.     The  converse  is  demonstrated 
as  prop.  19.   1.  EI.    Q.  E.  D. 

PROP.  IX.  FIG.  6. 
In  any  spherical  triangle  ABC,  if  the  sum  of  the 
sides  AB,  BC,  be  greater,  equal,  or  less  than  a  semi- 
circle, the  internal  angle  at  the  base  AC  will  be  greater, 
equal,  or  less  than  the  external  or  opposite  BCD;  and 
therefore  the  sum  of  the  angles  A  and  ACB  will  be 
greater,  equal,  or  less,  than  two  right  angles. 

Let  AC,  AB  produced  meet  in  D. 

1.  If  AB,  BC  be  equal  to  a  semicircle,  that  is,  to  AD,  BC,  BD 
Avill  be  equal,  that  is  (4.  of  this),  the  angle  D,  or  the  angle  A  will 
be  equal  to  the  angle  BCD. 

2.  If  AB,  BC  together  be  greater  than  a  semicircle,  that  is, 
greater  than  ABD,  BC  will  be  greater  than  BD;  and  therefore 
(8.  of  this)  the  angle  D,  that  is,  the  angle  A,  is  greater  than  the 
angle  BCD. 

3.  In  the  same  manner  it  is  shown,  that  if  AB,  BC  together  be 
less  than  a  semicircle,  the  angle  A  is  less  than  the  angle  BCD* 
And  since  the  angles  BCD,  BCA  are  equal  to  two  right  angles, 
if  the  angle  A  be  greater  than  BCD,  A  and  ACB  together  will 
be  greater  than  two  right  angles.  If  A  be  equal  to  BCD,  A  and 
ACB  together  will  be  equal  to  two  right  angles;  and  if  A  be 
less  than  BCD,  A  and  ACB  will  be  less  than  two  right  angles. 
Q.  E.  D. 

PROP.  X.    FIG.  7. 

If  the  angular  points.  A,  B,  C,  of  the  spherical  tri- 
angle ABC  be  the  poles  of  three  great  circles,  these 
great  circles  by  their  intersections  will  form  another 
triangle  FDE,  which  is  called  supplemental  to  the  for- 
mer; that  is,  the  sides  FD,  DE,  EF  are  the  supple- 
ments of  the  measures  of  the  opposite  angles  C,  B,  A 
of  the  triangle  ABC,  and  the  measures  of  the  angles 
F,  D,  E  of  the  triangle  FDE,  will  be  the  supplements  of 
the  sides,  AC,  BC,  BA  in  the  triangle  ABC. 

Let  AB  produced  meet  DE,  EF  in  G,  M,  and  AC  meet  FD, 
FE  in  K,  L,  and  BC  meet  FD,  DE  in  N,  H. 

Since  A  is  the  pole  of  FE,  and  the  circle  AC  passes  through  A, 
EF  will  pass  through  the  pole  of  AC  (13,  15.  1.  Th.),  and  since 
AC  passes  through  C,  the  pole  of  FD,  FD  will  pass  through  the 
pole  of  AC;  therefore  the  pole  of  AC  is  in  the  point  F,  in  which 
the  arches  DF,  EF  intersect  each  other.  In  the  same  manner,  D 
is  the  pole  of  BC,  and  E  the  pole  of  AB. 

And  since  F,  E  are  the  poles  of  AL,  AM,  FL  and  EM  are 
quadrants,  and  FL,  EM  together,  that  is,  FE  and  ML  together, 
3  E 


402  SPHERICAL  TRIGONOMETRY. 

are  equal  to  a  semicircle.  But  since  A  is  the  pole  of  ML,  ML  is 
the  measure  of  the  angle  BAG,  consequently  FE  is  the  supple- 
ment of  the  measure  of  the  angle  BAG.  In  the  same  manner, 
ED,  DF  are  the  supplements  of  the  measures  of  the  angles  ABC, 
BGA. 

Since  likewise  GN,  BH  are  quadrants,  GN,  BH  together,  that 
is  NH,  BG  together  are  equal  to  a  semicircle;  and  since  D  is  the 
pole  of  NH,  NH  is  the  measure  of  the  angle  FDE,  therefore  the 
measure  of  the  angle  FDE  is  the  supplement  of  the  side  BG.  In 
the  same  manner,  it  is  shown  that  the  measures  of  the  angles  DEF, 
EFD  are  the  supplements  of  the  sides  AB,  AG  in  the  triangle 
ABG.     Q.  E.  D. 

PROP.  XI.    FIG.  7. 

The  three  angles  of  a  spherical  triangle  are  greater 
than  two  right  angles,  and  less  than  six  right  angles. 

The  measures  of  the  angles  A,  B,  G,  in  the  triangle  ABG,  to- 
gether with  the  three  sides  of  the  supplemental  triangle  DEF, 
are  (10.  of  this)  equal  to  three  semicircles;  but  the  three  sides  of 
the  triangle  FDE,  are  (7.  of  this)  less  than  two  semicircles;  there-^ 
fore  the  measures  of  the  angles  A,  B,  G  are  greater  than  a  semi- 
circle; and  hence  the  angles  A,  B,  G  are  greater  than  two  right 
angles. 

AH  the  external  and  internal  angles  of  any  triangle  are  equal  to 
six  right  angles;  therefore  all  the  internal  angles  are  less  than 
six  right  angles. 

PROP.  XII.   FIG.  8. 

If  from  any  point  C,  which  is  not  the  pole  of  the 
great  circle  ABD,  there  be  drawn  arches  of  great  cir- 
cles CA,  CD,  CE,  CF,  &c.,  the  greatest  of  these  is  CA, 
which  passes  through  H  the  pole  of  ABD,  and  CB  the 
remainder  of  ACB  is  the  least,  and  of  any  others  CD, 
CE,  CF,  6lc.  CD,  which  is  nearer  to  CA,  is  greater 
than  CE,  which  is  more  remote. 

Let  the  common  section  of  the  planes  of  the  great  circles  AGB, 
ADB  be  AB;  and  from  G,  draw  GG  perpendicular  to  AB,  which 
will  also  be  perpendicular  to  the  plane  ADB  (4.  def.  11.);  join 
GD,  GE,  GF,  GD,  GE,  GF,  GA,  GB. 

Of  all  the  straight  lines  drawn  from  G  to  the  circumference 
ADB,  GA  is  the  greatest,  and  GB  the  least  (7.  3.);  and  GD, 
which  is  nearer  to  GA,  is  greater  than  GE,  which  is  more  remote. 
The  triangles  GGA,  GGD  are  right  angled  at  G,  and  they  have 
the  common  side  GG:  therefore  the  squares  of  GG,  GA  together, 
that  is,  the  square  of  GA,  is  greater  than  the  squares  of  GG,  GD 
together,  that  is,  the  square  of  GD;  and  GA  is  greater  than  GD, 
and  therefore  the  arch  GA  is  greater  than  GD.  In  the  same 
manner,  since  GD  is  greater  than  GE,  and  GE  than  GF,  &c.  it  is 
shown  that  GD  is  greater  than  GE,  and  GE  than  GF,  &c.;  and 


SPHERICAL  TRIGONOMETRY.  403 

consequently  the  arch  CD  greater  than  the  arch  CE;  and  the 
arch  CE  greater  than  the  arch  CF,  8cc.  And  since  GA  is  the 
greatest,  and  GB  the  least  of  all  the  straight  lines  drawn  from  G 
to  the  circumference  ADB,  it  is  manifest  that  CA  is  the  great- 
est, and  CB  the  least  of  all  the  straight  lines  drawn  from  C  to  the 
circumference:  and  therefore  the  arch  CA  is  the  greatest,  and 
CB  the  least  of  all  the  circles  drawn  through  C,  meeting  ADB. 
Q.  E.  D. 

PROP.  XIII.   FIG.  9. 

In  a  right  angled  spherical  triangle,  the  sides  are  of 
the  same  affection  with  the  opposite  angles;  that  is, 
if  the  sides  be  greater  or  less  than  quadrants,  the  op- 
posite angles  will  be  greater  or  less  than  right  angles. 

Let  ABC  be  a  spherical  triangle  right  angled  at  A;  any  side 

AB,  will  be  of  the  same  affection  with  the  opposite  angle  ACB. 
Case  1.  Let  AB  be  less  than  a  quadrant,  let  AE  be  a  quadrant, 

and  let  EC  be  a  great  circle  passing  through  E,  C.  Since  A  is  a 
right  angle,  and  AE  a  quadrant,  E  is  the  pole  of  the  great  circle 

AC,  and  ECA  a  right  angle:  but  ECA  is  greater  than  BCA, 
therefore  BCA  is  less  than  a  right  angle.    Q.  E.  D. 

Fig.  10. — Case  2.  Let  AB  be  greater  than  a  quadrant,  make 
AE  a  quadrant,  and  let  a  great  circle  pass  through  C,  E;  ECA  is 
a  right  angle  as  before,  and  BCA  is  greater  than  ECA,  that  is, 
greater  than  a  right  angle.    Q.  E.  D. 

PROP.  XIV. 

If  the  two  sides  of  a  right  angled  spherical  triangle 
be  of  the  same  affection,  the  hypothenuse  will  be  less 
than  a  quadrant:  and  if  they  be  of  different  affection, 
the  hypothenuse  will  be  greater  than  a  quadrant. 

Let  ABC  be  a  right  angled  spherical  triangle;  if  the  two  sides 
AB,  AC  be  of  the  same  or  of  different  affection,  the  hypothenuse 
BC  will  be  less  or  greater  than  a  quadrant. 

Fig.  9. — Case  I.  Let  AB,  AC  be  each  less  than  a  quadrant. 
Let  AE,  AG  be  quadrants;  G  will  be  the  pole  of  AB,  and  E  the 
pole  of  AC,  and  EG  a  quadrant;  but  by  prop.  12,  CE  is  greater 
than  CB,  since  CB  is  farther  off  from  CGD  than  CE.  In  the 
same  manner,  it  is  shown  that  CB,  in  the  triangle  CBD,  where 
the  two  sides  CD,  BD  are  each  greater  than  a  quadrant,  is  less 
than  CE,  that  is,  less  than  a  quadrant.    Q.  E.  D. 

Fig.  10. — Case.  2.  Let  AC  be  less,  and  AB  greater  than  a 
quadrant;  then  the  hypothenuse  BC  will  be  greater  than  a  quad- 
rant; for  let  AE  be  a  quadrant,  then  E  is  the  pole  of  AC,  and  EC 
will  be  a  quadrant.  But  CB  is  greater  than  CE  by  prop.  12, 
since  AC  passes  through  the  pole  of  ABD,    Q.  E.  D. 


404  SPHERICAL   TRIGONOMETRY. 

PROP.  XV. 

If  the  hypothenuse  of  a  right  angled  triangle  be 
greater  or  less  than  a  quadrant,  the  sides'  will  be  of 
different  or  the  same  aifection. 

This  is  the  converse  of  the  preceding,  and  demonstrated  in  the 
same  manner. 

PROP.  XVI. 

In  any  spherical  triangle  ABC,  if  the  perpendicular 
AD  from  A  on  the  base  BC,  fall  within  the  triangle,  the 
angles  B  and  C  at  the  base  will  be  of  the  same  affection; 
and  if  the  perpendicular  fall  without  the  triangle,  the 
angles  B  and  C  will  be  of  different  affection. 

Fig.  11. —  1.  Let  AD  fall  within  the  triangle;  then  (13.  of  this), 
since  ADB,  ADC  are  right  angled  spherical  triangles,  the  angles 
B,  C,  must  each  be  of  the  same  affection  as  AD. 

Fig.  12. — 2.  Let  AD  fall  without  the  triangle;  then  (13.  of  this), 
the  angle  B  is  of  the  same  affection  as  AD;  and  by  the  same  the 
angle  ACD  is  of  the  same  affection  as  AD;  therefore  the  angles 
ACB  and  AD  are  of  different  affection,  and  the  angles  B  and  ACB 
of  different  affection. 

Cor.  Hence  if  the  angles  B  and  C  be  of  the  same  affection,  the 
perpendicular  will  fall  within  the  base;  for,  if  it  did  not  ( 1 6.  of  this), 
B  and  C  would  be  of  different  affection.  And  if  the  angles  B  and 
C  be  of  opposite  affection,  the  perpendicular  will  fall  without  the 
triangle;  for,  if  it  did  not  (16.  of  this),  the  angles  B  and  C  would 
be  of  the  same  affection,  contrary  to  the  supposition. 

PROP.  XVn.    FIG.  13. 

In  right  angled  spherical  triangles,  the  sine  of  either 
of  the  sides  about  the  right  angle,  is  to  the  radius  of  the 
sphere,  as  the  tangent  of  the  remaining  side  is  to  the 
tangent  of  the  angle  opposite  to  that  side. 

Let  ABC  be  a  triangle,  having  the  right  angle  at  A;  and  let  AB 
be  either  of  the  sides;  the  sine  of  the  side  AB  will  be  to  the  radius, 
as  the  tangent  of  the  other  side  AC  to  the  tangent  of  the  angle 
ABC,  opposite  to  AC.  Let  D  be  the  centre  of  the  sphere;  join 
AD,  BD,  CD,  and  let  AE  be  drawn  perpendicular  to  BD,  which 
therefore  Avill  be  the  sine  of  the  arch  AB,  and  from  the  point  E,  let 
there  be  drawn  in  the  plane  BDC  the  straight  line  EF  at  right  an- 
gles to  BD,  meeting  DC  in  F,  and  let  AF  be  joined.  Since  there- 
fore the  straight  line  DE  is  at  right  angles  to  both  EA  and  EF,  it 
will  also  be  at  right  angles  to  the  plane  AEF  (4.  1 1.);  wherefore 
the  plane  ABD,  which  passes  through  DE,  is  perpendicular  to  the 
plane  AEF  (18.  11.),  and  the  plane  AEF  perpendicular  to  ABD; 
the  plane  ACD  or  AFD  is  also  perpendicular  to  the  same  ABD: 


.# 


SPHERICAL    TRIGONOMETRY.  405 

therefore  the  common  section,  viz.  the  straight  line  AF,  is  at  right 
angles  to  the  plane  ABD  (19.  1 1.),  and  FAE,  FAD  are  right  an- 
gles (3.  clef.  1 1.);  therefore  AF  is  the  tangent  of  the  arch  AC^  and 
in  the  rectilineal  triangle  AEF,  having  a  right  angle  at  A,  AE 
will  be  to  the  radius  as  AF  to  the  tangent  of  the  angle  AEF  (l.  PI. 
Tr.);  but  AE  is  the  sine  of  the  arch  AB,  and  AF  the  tangent  of 
the  arch  AC,  and  the  angle  AEF  is  the  inclination  of  the  planes 
CBD,  ABD  (6.  def.  ll.),  or  the  spherical  angle  ABC:  therefore 
the  sine  of  the  arch  AB  is  to  the  radius,  as  the  tangent  of  the  arch 
AC,  to  the  tangent  of  the  opposite  angle  ABC. 

Cor.  1.  If  therefore  of  the  two  sides,  and  an  angle  opposite  to 
one  of  them,  any  two  be  given,  the  third  will  also  be  given. 

Cor.  2.  And  since  by  this  proposition  the  sine  of  the  side  AB  is 
to  the  radius,  as  the  tangent  of  the  other  side  AC  to  the  tangent  of 
the  angle  ABC  opposite  to  that  side;  and  as  the  radius  is  to  the 
co-tangent  of  the  angle  ABC,  so  is  the  tangent  of  the  same  angle 
ABC  to  the  radius  (Cor.  2.  def.  PI.  Tr.);  by  equality,  the  sine  of 
the  side  AB  is  to  the  co-tangent  of  the  angle  ABC  adjacent  to  it, 
as  the  tangent  of  the  other  side  AC  to  the  radius. 

PROP.  XVIII.     FIG.   13. 

In  right  angled  spherical  triangles,  the  sine  of  the  hy- 
pothenuse  is  to  the  radius,  as  the  sine  of  either  side  is 
to  the  sine  of  the  angle  opposite  to  that  side. 

Let  the  triangle  ABC  be  right  angled  at  A,  and  let  AC  be 
either  of  the  sides;  the  sine  of  the  hypothenuse  BC  will  be  to  the 
radius,  as  the  sine  of  the  arch  AC  is  to  the  sine  of  the  angle 
ABC. 

Let  D  be  the  centre  of  the  sphere,  and  let  CG  be  drawn  perpen- 
dicular to  DB,  which  will  therefore  be  the  sine  of  the  hypothe- 
nuse BC;  and  from  the  point  G  let  there  be  drawn  in  the  plane 
ABD  the  straight  line  GH  perpendicular  to  DB,  and  let  CH  be 
joined:  CH  will  be  at  right  angles  to  the  plane  ABD,  as  was  shown 
in  the  preceding  proposition  of  the  straight  line  FA;  wherefore 
CHD,  CHG  are  right  angles,  and  CH  is  the  sine  of  the  arch  AC; 
and  in  the  triangle  CHG,  having  the  right  angle  CHG,  CG  is  to  the 
radius,  as  CH  to  the  sine  of  the  angle  CGH  (l.  PI.  Tr.);  but  since 
CG,  HG  are  at  right  angles  to  DGB,  which  is  the  common  section 
of  the  planes  CBD,  ABD,  the  angle  CGH  will  be  equal  to  the  in- 
clination of  these  planes  (6.  def.  1 1.);  that  is  to  the  spherical  angle 
ABC.  The  sine  therefore  of  the  hypothenuse  CB  is  to  the  radius, 
as  the  sine  of  the  side  AC  is  to  the  sine  of  the  opposite  angle 
ABC.     Q.  E.  D. 

CoR.  Of  these  three,  viz.  the  hypothenuse,  a  side,  and  the  angle 
opposite  to  that  side,  any  two  being  given,  the  third  is  also  given 
by  prop.  2. 

PROP.  XIX.     FIG.   14. 

In  right  angled  spherical  triangles,  the  co-sine  of  the 
hypothenuse  is  to  the  radius,  as  the  co-tangent  of  either 
of  the  angles  is  to  the  tangent  of  the  remaining  angle. 


406  SPHERICAL   TRIGONOMETRY. 

Let  ABC  be  a  spherical  triangle,  having  a  right  angle  at  A; 
the  co-sine  of  the  hypothenuse  BC  will  be  to  the  radius,  as  the  co- 
tangent of  the  angle  ABC  to  the  tangent  of  the  angle  ACB. 

Describe  the  circle  DE,  of  which  B  is  the  pole,  and  let  it  meet 
AC  in  F,  and  the  circle  BC  in  E;  and  since  the  circle  BD  passes 
through  the  pole  B  of  the  circle  DF,  DF  will  also  pass  through 
the  pole  BD  (13.  18.  1.  Theod.  sph.).  And  since  AC  is  perpen- 
dicular to  BD,  AC  will  also  pass  through  the  pole  of  BD5 
wherefore  the  pole  of  the  circle  BD  will  be  found  in  the  point 
where  the  circles  AC,  DE  meet,  that  is,  in  the  point  F:  the  arches 
FA,  FD  are  therefore  quadrants,  and  likewise  the  arches  BD,  BE: 
in  the  triangle  CEF,  right  angled  at  the  point  E,  CE  is  the  com- 
plement of  the  hypothenuse  BC  of  the  triangle  ABC,  EF  is  the 
complement  of  the  arch  ED,  which  is  the  measure  of  the  angle 
ABC,  and  FC  the  hypothenuse  of  the  triangle  CEF,  is  the  com- 
plement of  AC;  and  the  arch  AD,  which  is  the  measure  of  the  an- 
gle CFE,  is  the  complement  of  AB. 

But  (17.  of  this)  in  the  triangle  CEF,  the  sine  of  the  side  CE  is 
to  the  radius,  as  the  tangent  of  the  other  side  is  to  the  tangent  of 
the  angle  ECF  opposite  to  it,  that  is,  in  the  triangle  ABC,  the  co- 
sine of  the  hypothenuse  BC  is  to  the  radius,  as  the  co-tangent  of 
the  angle  ABC  is  to  the  tangent  of  the  angle  ACB.     Q.  E.  D. 

Cor.  1.  Of  these  three,  viz.  the  hypothenuse  and  the  two  angles, 
any  two  being  given,  the  third  will  also  be  given. 

Cor.  2.  And  since  by  this  proposition  the  co-sine  of  the  hypo- 
thenuse BC  is  to  the  radius  as  the  co-tangent  of  the  angle  ABC  to 
the  tangent  of  the  angle  ACB;  but  as  the  radius  is  to  the  co-tan- 
gent of  the  angle  ACB,  so  is  the  tangent  of  the  same  to  the  radius 
(Cor.  2.  def.  PI.  Tr.);  and  ex  aequo,  the  co-sine  of  the  hypothenuse 
BC  is  to  the  co-tangent  of  the  angle  ACB,  as  the  co-tangent  of  the 
angle  ABC  to  the  radius. 

PROP.  XX.     FIG.  14. 

In  right  angled  spherical  triangles,  the  co-sine  of  an 
angle  is  to  the  radius,  as  the  tangent  of  the  side  adja- 
cent to  that  angle  is  to  the  tangent  of  the  hypothenuse. 

The  same  construction  remaining;  in  the  triangle  CEF  (17.  of 
this),  the  sine  of  the  side  EF  is  to  the  radius,  as  the  tangent  of  the 
other  side  CE  is  to  the  tangent  of  the  angle  CFE  opposite  to  it; 
that  is,  in  the  triangle  ABC,  the  co-sine  of  the  angle  ABC  is  to 
the  radius,  as  (the  co-tangent  of  the  hypothenuse  BC  to  the  co-tan- 
gent of  the  side  AB,  adjacent  to  ABC,  or  as)  the  tangent  of  the 
side  AB  to  the  tangent  of  the  hypothenuse,  since  the  tangents  of 
two  arches  are  reciprocally  proportional  to  their  co-tangents. 
(Cor.  l.def.  PI.  Tr.) 

Cor.  And  since  by  this  proposition  the  co-sine  of  the  angle 
ABC  is  to  the  radius,  as  the  tangent  of  the  side  AB  is  to  the  tan- 
gent of  the  hypothenuse  BC;  and  as  the  radius  is  to  the  co-tangent 
of  BC,  so  is  the  tangent  of  BC  to  the  radius;  by  equality,  the  co- 
sine of  the  angle  ABC  will  be  to  the  co-tangent  of  the  hypothe- 
nuse BC,  as  the  tangent  of  the  side  AB,  adjacent  to  the  angle 
ABC,  to  the  radius. 


* 


SPHERICAL  TRIGONOMETRY.  407 


PROP.  XXI.    FIG.  14. 


In  right  angled  spherical  triangles,  the  co-sine  of 
either  of  the  sides  is  to  the  radius,  as  the  co-sine  of  the 
hypothenuse  is  to  the  co-sine  of  the  other  side. 

The  same  construction  remaining;  in  the  triangle  CEF,  the  sine 
of  the  hypothenuse  CF  is  to  the  radius,  as  the  sine  of  the  side  CE 
to  the  sine  of  the  opposite  angle  CFE  (18.  of  this);  that  is,  in  the 
triangle  ABC,  the  co-sine  of  the  side  CA  is  to  the  radius,  as  the 
co-sine  of  the  hypothenuse  BC  to  the  co-sine  of  the  other  side  BA. 
Q.  E.  D. 

PROP.  XXII.   FIG.  14. 

In  right  angled  spherical  triangles,  the  co-sine  of 
either  of  the  sides  is  to  the  radius,  as  the  co-sine  of  the 
angle  opposite  to  that  side  is  to  the  sine  of  the  other 
angle. 

The  same  construction  remaining;  in  the  triangle  CEF,  the 
sine  of  the  hypothenuse  CF  is  to  the  radius,  as  the  sine  of  the  side 
EF  is  to  the  sine  of  the  angle  ECF  opposite  to  it;  that  is,  in  the 
triangle  ABC,  the  co-sine  of  the  side  CA  is  to  the  radius,  as  the 
co-sine  of  the  angle  ABC  opposite  to  it,  is  to  the  sine  of  the  other 
angle.    Q.  E.  D. 


408  SPHERICAL  TRIGONOMETRY. 


OF  THE  CIRCULAR  PARTS. 

Fig.  15. — In  any  right  angled  spherical  triangle  ABC,  the  com- 
plement of  the  hypothenuse,  the  complements  of  the  angles,  and 
the  two  sides,  are  called  The  circular  parts  of  the  triangle^  as  if  it 
were  following  each  other  in  a  circular  order,  from  whatever  part 
we  begin:  thus,  if  we  begin  at  the  complement  of  the  hypothenuse, 
and  proceed  towards  the  side  BA,  the  parts  following  in  order 
will  be  the  complement  of  the  hypothenuse,  the  complement  of  the 
angle  B,  the  side  BA,  the  side  AC,  (for  the  right  angle  at  A  is  not 
reckoned  among  the  parts,)  and,  lastly,  the  complement  of  the 
angle  C.  And  thus  at  whatever  part  we  begin,  if  any  three  of 
these  five  be  taken,  they  either  will  be  all  contiguous  or  adjacent, 
or  one  of  them  will  not  be  contiguous  to  either  of  the  other  two: 
in  the  first  case,  the  part  which  is  between  the  other  two  is  called 
the  Middle  part,  and  the  other  two  are  called  Mjacent  extremes. 
In  the  second  case,  the  part  which  is  not  contiguous  to  either  of 
the  other  two  is  called  the  Middle  part,  and  the  other  two.  Oppo- 
site extremes.  For  example,  if  the  three  parts  be  the  complement 
of  the  hypothenuse  BC,  the  complement  of  the  angle  B,  and  the 
side  BAj  since  these  three  are  contiguous  to  each  other,  the  com- 
plement of  the  angle  B  will  be  the  middle  part,  and  the  comple- 
ment of  the  hypothenuse  BC  and  the  side  BA  will  be  adjacent  ex- 
tremes: but  if  the  complement  of  the  hypothenuse  BC,  and  the 
sides  BA,  AC  be  taken^  since  the  complement  of  the  hypothenuse 
is  not  adjacent  to  either  of  the  sides,  viz.  on  account  of  the  com- 
plements of  the  two  angles  B  and  C  intervening  between  it  and 
the  sides,  the  complement  of  the  hypothenuse  BC  will  be  the  mid- 
dle part,  and  the  sides  BA,  AC  opposite  extremes.  The  most 
acute  and  ingenious  Baron  Napier,  the  inventor  of  Logarithms, 
contrived  the  two  following  rules  concerning  these  parts,  by  means 
of  which  all  the  cases  of  right  angled  spherical  triangles  are  re- 
solved with  the  greatest  ease. 

RULE  I. 

The  rectangle  contained  by  the  radius  and  the  sine  of  the  middle 
part,  is  equal  to  the  rectangle  contained  by  the  tangents  of  the 
adjacent  parts. 

RULE  II. 

The  rectangle  contained  by  the  radius  and  the  sine  of  the  middle 
part,  is  equal  to  the  rectangle  contained  by  the  co-sines  of  the 
opposite  parts. 

These  rules  are  demonstrated  in  the  following  manner: 
Fig.  15.  First.  Let  either  of  the  sides,  as  BA,  be  the  middle 
part,  and  therefore  the  complement  of  the  angle  B,  and  the  side 
AC  will  be  adjacent  extremes.  And  by  Cor.  2.  prop.  17,  of  this, 
S,  BA  is  to  the  Co-T,  B,  as  T,  AC  is  to  the  radius,  and  therefore 
R  X  S,  BA=Co-T,  B  X  T,  AC. 

The  same  side  BA,  being  the  middle  part,  the  complement  of 
the  hypothenuse,  and  the  complement  of  the  angle  C,  are  opposite 


SPHERICAL  TRIGONTOMETRY.   '  409 

extremes;  and  by  prop.  18.  S,  BC  is  to  the  radius,  as  S,  BA  to  S, 
C;  therefore  R  X  S,  BA=S,  BC  X  S,  C. 

Secondly,  Let  the  complement  of  one  of  tlie  angles,  as  B,l)e  the 
middle  part,  and  the  complement  of  the  hypothenuse,  and  the  side 
BA  will  be  adjacent  extremes:  and  by  Cor.  prop.  30.  Co-S,  B  is 
to  Co-T,  BC,  as  T,  BA,  is  to  the  radius,  and  therefore  R  x  Co-S, 
B=Co-T,  BC  X  T,  BA. 

Again,  Let  the  complement  of  the  angle  B  be  the  middle  part, 
and  the  complement  of  the  angle  C,  and  the  side  AC  will  be  op- 
posite extremes:  and  by  prop.  22.  Co-S,  AC  is  to  the  radius,  as 
Co-S,  B  is  to  S,  C:  and  therefore  R  X  Co-S,  B=Co-S,  AC  x  S,  C. 

Thirdly,  Let  the  complement  of  the  hypothenuse  be  the  middle 
part,  and  the  complements  of  the  angles  B,  C,  will  be  adjacent  ex- 
tremes: but  by  Cor.  2.  prop.  19,  Co-S,  BC  is  to  Co-T,  B,  as  Co-T, 
C  to  the  radius:  therefore  R  X  Co-S,  BC=:Co-T,  C  X  Co-T,  A. 

Again,  Let  the  complement  of  the  hypothenuse  be  the  middle 
part,  and  the  sides  AB,  AC,  will  be  opposite  extremes:  but  by- 
prop.  21.  Co-S,  AC  is  to  the  radius,  as  Co-S,  BC,  to  Co-S,  BA> 
therefore  R  X  Co-S,  BC=Co-S,  BA  x  Co-S,  AC.    Q.  E.  D. 


3F 


410 


SPHERICAL  TRIGONOMETRY. 


SOLUTION    OF   THE    SIXTEEN    CASES    OF   RIGHT   ANGLED    SPHERICAL 

TRIANGLES. 


GENERAL  PROPOSITION. 

In  a  right  angled  spherical  triangle,  of  the  three 
sides  and  three  angles,  any  two  being  given,  besides 
the  right  angle,  the  other  three  may  be  found. 

In  the  following  Table  the  solutions  are  derived  from  the  pre- 
ceding propositions.  It  is  obvious  that  the  same  solutions  may 
be  derived  from  Baron  Napier's  tv^^o  rules  above  demonstrated, 
which,  as  they  are  easily  remembered,  are  commonly  used  in 
practice. 


Case 
1 
2 
3 

4 

Given. 

So't. 

AC,  C 

B 

R  :  Co-S,  AC  :  :  S,  C  :  Co-S,  B  :  and  B  is 
of  the  same  species  with  CA,  by  22,  and  13. 

AC,  B 

C 

Co-S,  AC  :  R  :  :  Co-S,  B  :  S,  C  :  by  22. 

B,  C 

AC 

S,  C  :  Co-S,  B  :  :  R  :  Co-S,  AC  :  by  22. 
and  AC  is  of  the  same  species  with  B.   13. 

BA,AC 

BC 

R  :  Co-S,  B A  :  :  Co-S,  AC  :  Co-S,  BC.  21. 
and  if  both  BA,  AC  be  greater  or  less  than 
a  quadrant,  BC  will  be  less  than  a  quadrant. 
But  if  they  be  of  different  affections,  BC 
will  be  greater  than  a  quadrant.   14. 

5 

BA,  BC 

AC 

Co-S,  BA  :  R  :  :  Co-S,  BC  :  Co-S,  AC.  21, 
and  if  BC  be  greater  or  less  than  a  quad- 
rant, BA,  AC  will  be  of  different  or  the 
same  affection:  by  15. 

6 

BA,AC 

B 

S,  BA  :  R  : :  T,  CA  :  T,  B,  17,  and  B  is 

of  the  same  affection  with  AC,  13. 

7 
8 

BA,  B 

AC 

R  :  S,  BA  :  :  T,  B  :  T,  AC.  17.     And  AC 

is  of  the  same  affection  with  B.  13. 

AC,  B 

BA 

T,  B  :  R  ::  T,CA  :  S,  BA.  17. 

SPHERICAL  TRIGONOMETRY. 


411 


Case 

Given. 

So't. 
AC 

9 

BC,  C 

R  :  Co-S,  C  :  ;  T,  BC  :  T,  CA.  20.    If  BC 

be  less  or  greater  than  a  quadrant,  C  and  B 
will  be  of  the  same  or  different  affection. 
15.  13. 

10 

AC,  C 

BC 

Co-S,  C  :  R  :  :  T,  AC  :  T,  BC.  20.     And 

BC  is  less  or  greater  than  a  quadrant,  ac- 
cording as  C  and  AC  or  C  and  B  are  of  the 
same  or  different  affection.   14.  I. 

11 

BC,  CA 

C 

T,BC  :  R  ::  T,  CA  :  Co-S,  C.  20.  IfBC 
be  less  or  greater  than  a  quadrant,  CA  and 
AB,  and  therefore  CA  and  C,  are  of  the 
same  or  different  affection.   15. 

12 

BC,  B 

AC 

R  :  S,  BC  :  :  S,  B  :  S,  AC.   18.     And  AC 

is  of  the  same  affection  with  B. 

13 

AC,  B 

BC 

S,  B  :  S,  AC  ::  R  :  S,  BC.   18. 

■ 

14 

BC,  AC 

B 

S,  BC  :  R  :  :  S,  AC  :  S,  B.   18.     And  B  is 

of  the  same  affection  with  AC. 

15 

B,  C 

BC 

T,C  :  R  :  :  Co-T,  B  :  Co-S,  BC.  19.  And 
according  as  the  angles  B  and  C  are  of  dif- 
ferent or  the  same  affection,  BC  will  be 
greater  or  less  than  a  quadrant.   14. 

16 

BC,  C 

B 

R  :  Co-S,  BC  :  :  T,  C  :  Co-T,  B.  19.  If  BC 

be  less  or  greater  than  a  quadrant,  C  and  B 
will  be  of  the  same  or  different  affection.  15. 

The  second,  eighth,  and  thirteenth  cases,  which  are  commonly 
called  ambiguous,  admit  of  two  solutions:  for  in  these  it  is  not 
determined  whether  the  side  or  measure  of  the  angle  sought  be 
greater  or  less  than  a  quadrant. 

PROP.  XXIII.     FIG.  16. 

In  spherical  triangles,  whether  right  angled  or  ob- 
lique angled,  the  sines  of  the  sides  are  proportional  to 
the  sines  of  the  angles  opposite  to  them. 

First,  Let  ABC  be  a  right  angled  triangle,  having  a  right  angle 
at  A;  therefore  by  prop.  18,  the  sine  of  the  hypothenuse  BC  is  to 
the  radius  (or  the  sine  of  the  right  angle  at  A)  as  the  sine  of  the 
side  AC  to  the  sine  of  the  angle  B.     And  in  like  manner,  the  sine 


412  SPHERICAL  TRIGONOMETRY. 

of  BC  is  to  the  sine  of  the  angle  A  as  the  sine  of  AB  to  the  sine 
of  the  angle  C^  wherefore  (11.  5.)  the  sine  of  the  side  AC  is  to 
the  sine  of  the  angle  B,  as  the  sine  of  AB  to  the  sine  of  the  an- 
gle C. 

Secondly,  Let  BCD  be  an  oblique  angled  triangle,  the  sine  of 
either  of  the  sides  BC,  will  be  to  the  sine  of  either  of  the  other 
two  CD,  as  the  sine  of  the  angle  D  opposite  to  BC  is  to  the  sine 
of  the  angle  B  opposite  to  the  side  CD.  Through  the  point  C, 
let  there  be  drawn  an  arch  of  a  great  circle  CA  perpendicular 
upon  BDj  and  in  the  right  angled  triangle  ABC  (18.  of  this)  the 
sine  of  JBC  is  to  the  radius,  as  the  sine  of  AC  to  the  sine  of  the 
angle  B;  and  in  the  triangle  ADC  (by  18.  of  this):  and,  by  inver- 
sion, the  radius  is  to  the  sine  of  DC  as  the  sine  of  the  angle  D  to 
the  sine  of  AC:  therefore,  ex  sequo  perturbate,  the  sine  of  BC  is 
to  the  sine  of  DC,  as  the  sine  of  the  angle  D  to  the  sine  of  the 
angle  B.    Q.  E.  D. 

PROP.  XXIV.    FIG.  17.  18. 

In  oblique  angled  spherical  triangles,  having  drawn 
a  perpendicular  arch  from  any  of  the  angles  upon  the 
opposite  side,  the  co-sines  of  the  angles  at  the  base  are 
proportional  to  the  sines  of  the  vertical  angles. 

Let  BCD  be  a  triangle,  and  the  arch  CA  perpendicular  to  the 
base  BD;  the  co-sine  of  the  angle  B  will  be  to  the  co-sine  of  the 
angle  D,  as  the  sine  of  the  ansjle  BCA  to  the  sine  of  the  angle 
PCA. 

For,  by  22,  the  co-sine  of  the  angle  B  is  to  the  sine  of  the  an- 
gle BCA  as  (the  co-sine  of  the  side  AC  is  to  the  radius;  that  is, 
by  prop.  22,  as  (  the  co-sine  of  the  angle  D  to  the  sine  of  the  an- 
gle DCA;  and,  by  permutation,  the  co-sine  of  the  angle  B  is  to 
the  co-sine  of  the  angle  D,  as  the  sine  of  the  angle  BCA  to  the 
sine  of  the  angle  DCA.     Q.  E.  D. 

PROP.  XXV.  FIG.  17.  18. 

The  same  things  remaining,  the  co-sines  of  the  sides 
BC,  CD,  are  proportional  to  the  co-sines  of  the  bases 
34,  AD. 

For,  by  21,  the  co-sine  of  BC  is  to  the  co-sine  of  BA,  as  (the 
co-sine  of  AC  to  the  radius;  that  is,  by  21,  as)  the  co-sine  of  CD 
is  to  the  co-sine  of  AD:  wherefore,  by  permutation,  the  co-sines 
of  the  sides  BC,  CD  are  proportional  to  the  co-sines  of  the  bases 
BA,  AD.    Q.  E.  D. 

PROP.  XXVI.  FIG.  17.  18. 

The  same  construction  remaining,  the  sines  of  the 
bases  BA,  AD  are  reciprocally  proportional  to  the  tan- 
gents of  the  angles  B  and  D  at  the  base. 


SPHERICAL   TRIGONOMETRY.  413 

For,  by  17,  the  sine  of  BA  is  to  the  radius,  as  the  tangent  of 
AC  to  the  tangent  of  the  angle  B;  and  by  17,  and  inversion,  the 
radius  is  to  the  sine  of  AD,  as  the  tangent  of  D  to  the  tangent  of 
AC:  therefore,  ex  aequo  perturbate,  the  sine  of  BA  is  to  the  sine 
of  AD,  as  the  tangent  of  D  to  the  tangent  of  B. 

PROP.   XXVII.    FIG.   17.   18. 

The  co-sines  of  the  vertical  angles  are  reciprocally 
proportional  to  the  tangents  of  the  sides. 

For,  by  prop.  20,  the  co-sine  of  the  angle  BCA  is  to  the  radius 
as  the  tangent  of  CA  is  to  the  tangent  of  BC;  and  by  the  same 
prop.  20,  and  by  inversion,  the  radius  is  to  the  co-sine  of  the  an- 
gle DCA,'  as  the  tangent  of  DC  to  the  tangent  of  CA:  therefore, 
ex  aequo  perturbate,  the  co-sine  of  the  angle  BCA  is  to  the  co- 
sine of  the  angle  DCA,  as  the  tangent  of  DC  is  to  the  tangent  of 
BC.     Q.  E.  D. 

LEMMA.    FIG.   19.  20. 

In  right  angled  plane  triangles,  the  hypothenuse  is  to 
the  radius,  as  the  excess  of  the  hypothenuse  above 
either  of  the  sides  to  the  versed  sine  of  the  acute  angle 
adjacent  to  that  side,  or  as  the  sum  of  the  hypothenuse, 
and  either  of  the  sides  to  the  versed  sine  of  the  exterior 
angle  of  the  triangle. 

Let  the  triangle  ABC  have  a  right  angle  at  B;  AC  will  be  to 
the  radius  as  the  excess  of  AC  above  AB,  to  the  versed  sine  of 
the  angle  A  adjacent  to  AB;  or  as  the  sum  of  AC,  AB  to  the 
versed  sine  of  the  exterior  angle  CAK. 

With  any  radius  DE,  let  a  circle  be  described,  and  from  D 
the  centre,  let  DF  be  drawn  to  the  circumference,  making  the 
angle  EDF  equal  to  the  angle  BAC,  and  from  the  point  F,  let 
FG  be  drawn  perpendicular  to  DE;  let  AH,  AK  be  made  equal 
to  AC,  and  DL  to  DE:  DG  therefore  is  the  co-sine  of  the  angle 
EDF  or  BAC,  and  GE  its  versed  sine:  and  because  of  the  equi- 
angular triangles  ACB,  DFG,  AC  or  AH  is  to  DF  or  DE,  as  AB 
to  DG:  therefore  (19.  5.)  AC  is  to  the  radius  DE  as  BH  to  GE, 
the  versed  sine  of  the  angle  EDF  or  BAC:  and  since  AH  is  to 
DE,  as  AB  to  DG  (12.  5.),  AH  or  AC  will  be  to  the  radius  DE 
as  KB  to  LG,  the  versed  sine  of  the  angle  LDF  or  KAC.  Q. 
E.  D. 

PROP.  XXVIII.     FIG.  21.  22. 

In  any  spherical  triangle,  the  rectangle  contained  by 
the  sines  of  two  sides,  is  to  the  square  of  the  radius, 
as  the  excess  of  the  versed  sines  of  the  third  side  or 
base,  and  the  arch,  which  is  the  excess  of  the  sides,  is 
to  the  versed  sine  of  the  angle  opposite  to  the  base. 


414  SPHERICAL  TRIGONOMETRY. 

Let  ABC  be  a  spherical  triangle;  the  rectangle  contained  by 
the  sines  of  AB,  BC  will  be  to  the  square  of  the  radius,  as  the 
excess  of  the  versed  sines  of  the  base  AC,  and  of  the  arch,  which 
is  the  excess  of  AB,  BC,  to  the  versed  sine  of  the  angle  ABC 
opposite  to  the  base. 

Let  D  be  the  centre  of  the  sphere,  and  let  AD,  BD,  CD  be 
joined,  and  let  the  sines  AE,  CF,  CG  of  the  ajxhes  AB,  BC,  AC 
be  drawn;  let  the  side  BC  be  greater  than  BA,  and  let  BH  be 
made  equal  to  BC;  AH  will  therefore  be  the  excess  of  the  sides 
BC,  BA;  let  HK  be  drawn  perpendicular  to  AD,  and  since  AG  is 
the  versed  sine  of  the  base  AC,  and  AK  the  versed  sine  of  the 
arch  AH,  KG  is  the  excess  of  the  versed  sines  of  the  base  AC, 
and  of  the  arch  AH,  which  is  the  excess  of  the  sides  BC,  BA:  let 
GL  likewise  be  drawn  parallel  to  KH,  and  let  it  meet  FH  in  L:  let 
CL,  DH  be  joined,  3,nd  let  AD,  FH  meet  each  other  in  M. 

Since  therefore  in  the  triangles  CDF,  HDF,  DC,  DH  are  equal, 
DF  is  common,  and  the  angle  FDC  equal  to  the  angle  FDH;  be- 
cause of  the  equal  arches  BC,  BH,  the  base  HF  will  be  equal  to 
the  base  FC,  and  the  angle  HFD  equal  to  the  right  angle  CFD: 
the  ^straight  line  DF  therefore  (4.  11.)  is  at  right  angles  to  the 
plane  CFH:  wherefore  the  plane  CFH  is  at  right  angles  to  the 
plane  BDH,  which  passes  through  DF  (18.  11.).  In  like  manner, 
since  DG  is  at  right  anf^les  to  both  GC  and  GL,  DG  will  be  per- 
pendicular to  the  plane  CGL;  therefore  the  plane  CGL  is  at  right 
angles  to  the  plane  BDH,  which  passes  through  DG:  and  it  was 
shown,  that  the  plane  CFH  or  CFL  was  perpendicular  to  the 
same  plane  BDH;  therefore  the  common  section  of  the  planes 
CFL,  CGL,  viz.  the  straight  line  CL  is  perpendicular  to  the  plane 
BDA  (19.  1 1.),  and  therefore  CLF  is  a  right  angle:  in  the  triangle 
CFL  having  the  right  angle  CLF,  by  the  lemma,  CF  is  to  the  ra- 
dius, as  LH,  the  excess,  viz.  of  CF  or  FH  above  FL,  is  to  the 
versed  sine  of  the  angle  CFL;  but  the  angle  CFL  is  the  inclina- 
tion of  the  planes  BCD,  BAD,  since  FC,  FL  are  draAvn  in  them  at 
right  angles  to  thecommon  section  BF:  the  spherical  angle  ABC 
is  therefore  the  same  with  the  angle  CFL;  and  therefore  CF  is  to 
the  radius  as  LH  to  the  versed  sine  of  the  spherical  angle  ABC; 
and  since  the  triangle  AED  is  equiangular  (to  the  triangle  MFD, 
and  therefore)  to  the  triangle  MGL,  AE  will  be  to  the  radius  of 
the  sphere  AD,  as  (MG  to  ML;  that  is,  because  of  the  parallels, 
as)  GK  to  LH:  the  ratio  therefore  which  is  compounded  of  the 
ratios  of  AE  to  the  radius,  and  of  CF  to  the  same  radius;  that  is, 
(23.  6.)  the  ratio  of  the  rectangle  contained  by  AE,  CF  to  the 
square  of  the  radius,  is  the  same  with  the  ratio  compounded  of 
the  ratio  of  GK  to  LH,  and  the  ratio  of  LH  to  the  versed  sine  of 
the  angle  ABC;  that  is,  the  same  with  the  ratio  of  GK  to  the 
versed  sine  of  the  angle  ABC;  therefore,  the  rectangle  contained 
by  AE,  CF,  the  sines  of  the  sides  AB,  BC,  is  to  the  square  of  the 
radius,  as  GK  the  excess  of  the  versed  sines  AG,  AK,  of  the  base 
AC,  and  the  arch  AH,  which  is  the  excess  of  the  sides,  to  the 
versed  sine  of  the  angle  ABC  opposite  to  the  base  AC.    Q.  E.  D. 


SPHERICAL  TRIGONOMETRY,  .  415 

PROP.  XXIX.    FIG.  23. 

The  rectangle  contained  by  half  of  the  radius,  and 
the  excess  of  the  versed  sines  of  two  arches,  is  equal  to 
the  rectangle  contained  by  the  sines  of  half  the  sum, 
and  half  the  difference  of  the  same  arches. 

Let  AB,  AC  be  any  two  arches,  and  let  AD  be  made  equal  to 
AC  the  lessj  the  arch  DB  therefore  is  the  sum,  and  the  arch  CB 
the  difference  of  AC,  AB:  through  E  the  centre  of  the  circle,  let 
there  be  drawn  a  diameter  DEF,  and  AE  joined,  and  CD  likewise 
perpendicular  to  it  in  Gj  and  let  BH  be  perpendicular  to  AE, 
and  AH  will  be  the  versed  sine  of  the  arch  AB,  and  AG  the  versed 
sine  of  AC,  and  HG  the  excess  of  these  versed  sines:  let  BD,  BC, 
BF  be  joined,  and  FC  also  meeting  BH  in  K. 

Since  therefore  BH,  CG  are  parallel,  the  alternate  angles  BKC, 
KCG  will  be  equal;  but  KCG  is  in  a  semicircle;  and  therefore  a 
right  angle;  therefore  BKC  is  a  right  angle;  and  in  the  triangles, 
DFB,  ci3K,  the  angles  FDB,  BCK,  in  the  same  segment  are  equal, 
and  FBD,  BKC  are  right  angles;  the  triangles  DFB,  CBK  are 
therefore  equiangular;  wherefore  DF  is  to  DB,  as  BC  to  CK,  or 
HG;  and  therefore  the  rectangle  contained  by  the  diameter  DF 
and  HG,  is  equal  to  that  contained  by  DB,  BC;  wherefore  the 
rectangle  contained  by  a  fourth  part  of  the  diameter,  and  HG,  is 
equal  to  that  contained  by  the  halves  of  DB,  BC:  but  half  the 
chord  DB  is  the  sine  of  half  the  arch  DAB,  that  is,  half  the  sum 
of  the  arches  AB,  AC;  and  half  the  chord  of  BC  is  the  sine  of 
half  the  arch  BC,  which  is  the  difference  of  AB,  AC.  Whence 
the  proposition  is  manifest. 

PROP.  XXX.    FIG.  19.  24. 

The  rectangle  contained  by  half  of  the  radius,  and 
the  versed  side  of  any  arch,  is  equal  to  the  square  of 
the  sine  of  half  the  same  arch. 

Let  AB  be  an  arch  of  a  circle,  C  its  centre,  and  AC,  CB,  BA 
being  joined;  let  AB  be  bisected  in  D,  and  let  CD  be  joined,  which 
will  be  perpendicular  to  BA,  and  bisect  it  in  E  (4.  1.),  BE  or  AE 
therefore  is  the  sine  of  the  arch  DB  or  AD,  the  half  of  AB:  let 
BF  be  perpendicular  to  AC,  and  AF  will  be  the  versed  sine  of  the 
arch  BA;  but,  because  of  the  similar  triangles  CAE,  BAF,  CA  is 
to  AE,  as  AB,  that  is,  twice  AE,  to  AF;  and  by  halving  the  ante- 
cedents, half  of  the  radius  CA  is  to  AE,  the  sine  of  the  arch  AD. 
as  the  same  AE,  to  AF  the  versed  sine  of  the  arch  AB.  Where- 
fore by  16,  6,  the  proposition  is  manifest. 

PROP.  XXXI.    FIG.  25. 

In  a  spherical  triangle,  the  rectangle  contained  by 
the  sines  of  the  two  sides,  is  to  the  square  of  the  radius, 
as  the  rectangle  contained  by  the  sine  of  the  arch  which 


416  SPHERICAL   TRIGONOMETRY. 

is  half  the  sum  of  the  base,  and  the  excess  of  the  sides, 
and  the  sine  of  the  arch,  which  is  half  the  difference  of 
the  same,  to  the  square  of  the  sine  of  half  the  angle 
opposite  to  the  base. 

Let  ABC  be  a  spherical  triangle,  of  which  the  two  sides  are 
AB,  BC,  and  base  AC,  and  let  the  less  side  BA  be  produced,  so 
that  BD  shall  be  equal  to  BC:  AD  therefore  is  the  excess  of  BC, 
BA,*  and  it  is  to  be  shown,  that  the  rectangle  contained  by  the 
sines  of  BC,  BA  is  to  the  square  of  the  radius,  as  the  rectangle 
contained  by  the  sine  of  half  the  sum  of  AC,  AD,  and  the  sine  of 
half  the  difference  of  the  same  AC,  AD  to  the  square  of  the  sine 
of  half  the  angle  ABC,  opposite  to  the  base  AC. 

Since  by  prop.  28,  the  rectangle  contained  by  the  sines  of  the 
sides  BC,  BA  is  to  the  square  of  the  radius,  as  the  excess  of  the 
versed  sines  of  the  base  AC  and  AD,  to  the  versed  sine  of  the  an- 
gle B;  that  is,  (l.  6.)  as  the  rectangle  contained  by  half  the  radius, 
and  that  excess,  to  the  rectangle  contained  by  half  the  radius,  and 
the  versed  sine  of  B;  therefore  (29.  30.  of  this)  the  rectangle  contain- 
ed by  the  sines  of  the  sides  BC,  B  A  is  to  the  square  of  the  radius,  as 
the  rectangle  contained  by  the  sine  of  the  arch,  which  is  half  the 
sum  of  AC,  AD,  and  the  sine  of  the  arch  which  is  half  the  differ- 
ence of  the  same  AC,  AD  is  to  the  square  of  the  sine  of  half  the 
angle  ABC.     Q.  E.  D. 


SPHERICAL  TRIGONOMETRY. 


417 


SOLUTION    OF    THE    TWELVE    CASES    Of    OBLIQUE    ANGLED    SPHERICAL 

TRIANGLES. 

GENERAL  PROPOSITION. 

In  an  oblique  angled  spherical  triangle,  of  the  three 
sides  and  three  angles,  any  three  being  given,  the  other 
three  may  be  found. 


Given. 

Sought. 

1 

B,  D, and 
BC,    two 

angles   & 
a  side  op- 
posite 
one  of 
them. 
Fig.  26, 
27. 

C. 

Co-S,  BC  :  R  :  ;  Co-T,  B  :  T,  BCA.  19. 

Likewise  by  24.  Co-S,  B  :  S,  BCA  : :  Co- 
S,  D  :  S,  DCAj  wherefore  BCD  is  the  sum 
or  difference  of  the  angles  DCA,  BCA  ac- 
cording as  the  perpendicular  CA  falls  with- 
in or  without  the  triangle  BCD;  that  is  (16. 
of  this,)  according  as  the  angles  B,  D  are  of 
the  same  or  different  affection. 

2 

B,  C,  and 
BC,    two 

angles  & 
the     side 
between 
them. 

D. 

Co-S,  BC  :  R  :  :  Co-T,  B  :  T,  BCA.  19. 

and  also  by  24.  S,  BCA  :  S,  DCA  :  :  Co- 
S,  B  :  Co-S,  D;  and  according  as  the  angle 
BCA  is  less  or  greater  than  BCD,  the  per- 
pendicular CA  falls  within  or  without  the 
triangle  BCD;  and  therefore  (16.  of  this,) 
the  angles  B,  D  will  be  of  the  same  or  dif- 
ferent affection. 

3 

BC,  CD, 

and  B. 

BD. 

R  :  Co-S,  B  :  :  T,  BC  :  T,  BA.  20.  and 
Co-S,  BC  :  Co-S,  BA  :  :  Co-S,  DC  :  Co-S, 
DA.  25.  and  BD  is  the  sum  or  difference  of 
BA,  DA. 

4 

BC,  DB, 

and  B. 

CD. 

R  :  Co-S,  B  ;  :  T,  BC  :  T,  BA.  20.  and 
Co-S,  BA  :  Co-S,  BC  :  :  Co-S,  DA  :  Co-S, 
DC.  25.  and  according  as  DA,  AC  are  of 
the  same  or  different  affection,  DC  will  be 
less  or  greater  than  a  quadrant.   14. 

5 

B,D,and 
BC. 

DB. 

R  :  Co-S,  B  :  :  T,  BC  :  T,  BA.  20.  and 
T,  D  :  T,  B  :  :  S,  B A  :  S,  DA.  26.^and  BD 
is  the  sum  or  difference  of  BA,  DA. 

3  G 


418 


SPHERICAL  TRIGONOMETRY. 


Given. 

Sought. 

6 

BC,    BD 

and  B. 

D. 

R  :  Co-S,  B  :  :  T,  BC  :  T,  BA.  20.  and 
S,  DA  :  S,  BA  :  :  T,  B  :  T,  Dj  and  ac- 
cording as  BD  is  greater  or  less  than  BA, 
the  angles  B,  D  are  of  the  same  or  different 
affection.   16. 

7 

BC,    DC 

and  B. 

C. 

Co-S,  BC  :  R  :  :  Co-T,  B  :  T,  BCA.  19. 
and  T,  DC  :  T,  BC  :  :  Co-S,  BCA  :  Co-S, 
DCA.  27.  the  sum  or  difference  of  the  an- 
gles BCA,  DCA  is  equal  to  the  angle  BCD. 

8 

B,  C  and 
BC. 

DC. 

Co-S,  BC  :  R  :  :  Co-T,  B  :  T,  BCA.  19. 
also  bv  27.  Co-S,  CDA  :  Co-S,  BCA  : :  T, 
BC  :  T,  DC.  27.  if  DCA  and  B  be  of  the 
same  affection;  that  is  (13.)  if  AD  and  CA 
be  similar,  DC  will  be  less  than  a  quadrant. 
14.  and  if  AD,  CA  be  not  of  the  same  affec- 
tion, DC  is  greater  than  a  quadrant.   14. 

9 

BC,    CD 

and  B. 

D. 

S,  CD  :  S,  B  :  :  S,  BC  :  S,  D. 

10 

B,  D  and 
BC. 

DC. 

S,  D  :  S,  BC  :  :  S,  B  :  S,  DC. 

11 

BC,  BA, 
AC. 

Fig.  25. 

B. 

S,  AB  X  S,  BC  :  R^  :  :  S,  AC+AD  x  S, 
AC— AD  :  Sq.  ABC.     See  Fig.  25.     AD 

2                                    2 

being  the  difference  of  the  sides  BC,  BA. 

12 

A,  B,  C. 

Fig.  7. 

The 
sides. 

See  Fig.  7. 
In  the  triangle  DEF,  DE,  EF,  FD  are  re- 
spectively the  supplements  of  the  measures 
of  the  given  angles  B,  A,  C,  in  the  triangle 
BAC;   the  sides  of  the  triangle  DEF  are 
therefore  given,  and  by  the  preceding  case 
the  angles  D,  E,  F  may  be  found,  and  the 
sides  BC,  BA,  AC  are  the  supplements  of 
the  measures  of  these  angles. 

SPHERICAL  TRIGONOMETRY.  419 

The  3d,  5thj  7th,  9th,  10th  cases,  -which  are  commonly  called 
ambiguous,  admit  of  two  solutions,  either  of  which-  will  answer 
the  conditions  required^  for,  in  these  cases,  the  measure  of  the 
angle  or  side  sought,  may  be  either  greater  or  less  than  a  quad- 
rant, and  the  two  solutions  will  be  supplements  to  each  other. 
(Cor.  to  def.  4.  6.  PL  Tr.) 

If  from  any  of  the  angles  of  an  oblique  angled  spherical  trian- 
gle, a  perpendicular  arch  be  drawn  upon  the  opposite  side,  most 
of  the  cases  of  oblique  angled  triangles  may  be  resolved  by  means 
of  Napier's  rules. 


FINIS. 


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